Good Question DSP

8/16/2019 Good Question DSP

1/36

TWO MARKS

Unit – I

1. What is a Signal?

A function of one or more independent variables which contain some

information called signal.

2. What is a system?

A system is a set of elements or functional block that are connected together and

produces an output in response to an input signal.

3. Define Piecewise continuous signals?

Piecewise continuous signals possess different expressions over different

intervals.

4.

Define continuous signals?Continuous signals are expressed by a signal expression for all time.

5. Define Periodic Signals?

Periodic signals are infinite duration signals that repeat the same pattern

endlessly. The smallest repetition interval is called the period T and leads to a formal

definition.

( ) ( )nTtxtx pp ±=

6.

Define energy signal and power signal?

Signal with finite energy is called an energy signal (or) square integrable.

Signal with finite power are called power signals.

7. Define ever symmetric and odd symmetric and give its expression in continuous

time signals

If a signal is identical to its folded version with ( ) ( )txtx −= , it is called even

symmetric.

( ) ( )txtx ee −=

If a signal and its folded version differs only in sign with x(t) = -x(-t), it is called

odd symmetric.

( ) ( )txtx oo −=


2/36

8. Define impulse function?

An impulse is a tall Narrow spike with finite area and Infinite energy

∫∞

∞−

=δ 1dc)x( ( )=δ t

9. What are the properties of the impulse

1. Scaling Property

( ) ( )t t δ α

α δ 11

1=

2. Product property

( ) ( ) )t()(xttx α−δ∞=α−δ

3. Sifting property

∫∞

∞−

α=α−δ )(xdt)t()t(x

10. Define doublet and give its three properties

The derivative of an impulse of an impulse is called a doublet denoted by )t('δ

Scaling property:

( ) ( )t1

t '' δ

αα

=αδ

Product property:

( ) ( ) ( ) ( ) ( ) ( )t0xt0xttx ''' δ−δ=δ

Sifting property:

( ) ( ) ( )0xdtttx '' −=δ∫α

α−

11. Define Superposition:

A linear operator obeys superposition:

Superposition principle:

aO{x1(t)} + bO{x2(t)} = O{a x1(t) + b x2(t)}

Superposition implies both homogeneity and additivity

Homogenity:

O{a x(t)} = aO{x(t)

0, t ≠ 0

α, t = 0


3/36

Additinity:

O{x1(t)} + O{x2(t)} = O{x1 (t) + x2(t)}

12. What makes a system Nonlinear?

1. Nonlinear elements

2. Nonzero initial conditions

3. Internal sources

13. What makes a system differential equation nonlinear or time varying

• Terms containing product of the i/p and/or o/p make a system equation

nonlinear. A constant term also makes a system equation.

• Cofficients of the i/p that are explicit functions of ‘t’ make a system

equation time varying. Time scaled i/p’s or o/p’s such as y(2t) also

make a system equation time varying.

14. What makes a system differential equation noncausal and static

• It is noncausal, if the o/p terms have the form y(t) and any i/p term

contains x(t+α), α>o.

• It is static if no derivatives are present, and every term in and y has

identical arguments

15.

Define foxed response?Forced response arises due to the interaction of the system with the i/p and thus

depends on both the i/p and system details. It satisfies the given differential equation

and has the same form as the i/p.

16.

Define total Response:

Total Response is found by first adding the forced and natural response and then

evaluating the undetermined constants using the prescribed initial conditions.

17. What are the methods to sketch x(αt - β)

Method 1:Method 1:Method 1:Method 1:

Shift right by β = [x(t) ⇒ x(t-β)]

Then compress by α = [x(t β) ⇒ x(αt - β)]


4/36

Method 2:Method 2:Method 2:Method 2:

Compress by α = [x(t) ⇒ x(αt)]

Then shift right byα

β = ( ) ( )

β−α=

α

β−α⇒α txtxtx

18. What is the resultant symmetry of the sum or product of two symmetric signals

1. xe (t) + ye(t) = Even symmetry

2. xo (t) + yo(t) = Odd symmetry

3. xe (t) + yo(t) = No symmetry

4. xe (t) ye(t) = Even symmetry

5. xo (t) yo(t) = Even symmetry

6. xe (t) yo(t) = Odd symmetry

19. How can we relate the impulse and a step function

,dt

)t(du)t( =δ ∫

∞−

δ=t

dt)t()t(u

20. Find the energy of the signal x(t) = 2e-t – 6e

-2t, t>0

Solution:

Ex = ∫∞

o

2 dt)t(x

= ( )dte36e24e4o

t4t3t2

∫

∞

−−− +−

= 2 – 8+9

= 3J


5/36

UNIT I

1) Briefly explain about Impulse and Doublet.

Impulse:Impulse is a tall narrow spike with finite area and infinite energy

∫ δ(τ) dτ =1 δ(τ) = {Three properties of impulse:

i) Scaling property

δ(αt) =

ii) Product property

x(t)δ(t –

α) = x(

α)δ(t –

α)

iii) Sifting property

∫ x(t)δ(t – α)dt = x(α)

Doublet:Derivative of impulse is called doublet and is denoted by δ΄(t)

δ΄(t) = { ∫ δ΄(t)dt = 0Properties of doublet:

i) Scaling property:

δ΄(αt) =

ii) Product Property

x(t)δ΄(t) = x(0) δ΄(t) - x΄(0) δ(t)

iii)

Sifting Property

∫ x(t)δ΄(t) dt = -x΄(0)2) Let y΄΄(t) + 3y΄(t) +2y(t) = x(t) with x(t) =4e

-3t and initial conditions y(0) = 3

and y΄(0) = 4. Find its zero-input response and zero-state response.

Solution:Characteristic equation: s

2+3s+2 = 0

Roots are: s1 = -1, s2 = -2

-∞ ∞ 0, t ≠ 0

∞, t = 0

|α |δ(t)1

-∞

∞

-∞ ∞ 0, t ≠ 0

Undefined, t = 0

|α |δ(t)1α ΄

-∞

∞


6/36

ZIR:Found from yN(t) and the prescribed initial conditions.

yzi(t) =K1e-t + K2e

-2t;

K2 = -7, K1 = 10

yzi(t) = 10e-t -7e

-2t

ZSR:

Found from general form of y(t) with zero initial conditions.x(t) = 4e

-3t ; So yF(t) = Ce

-3t

C = 2 and yF(t) = 2e-3t

yzs((t) = K1e-t + K2e

-2t +2e

-3t

with initial conditions yzs(0) = 0 and yzs΄(0) = 0, K2= -4 and K1 = 2

yzs(t) = 2e-t – 4e-2t + 2e-3t

3) Find the response y΄΄(t) + 3y΄(t) +2y(t) = 2x΄(t) + x(t), with x(t) = 4e-3t

, y(0) =

0 and y΄(0) = 1

Solution:Characteristic equation: s

2+3s+2 = 0

Roots are: s1 = -1, s2 = -2

ZIR:Found from yN(t) and the prescribed initial conditions.

yzi(t) =K1e-t + K2e

-2t;

K2 = -1, K1 = 1

yzi(t) = e-t -e

-2t

To find yo(t):

Found from general form of y(t) with zero initial conditions.

x(t) = 4e-3t ; So yF(t) = Ce-3t

C = 2 and yF(t) = 2e-3t

Yo((t) = K1e-t + K2e

-2t +2e

-3t

with initial conditions yzs(0) = 0 and yzs΄(0) = 0, K2= -4 and K1 = 2

yo(t) = 2e-t – 4e-

2t + 2e

-3t

To find ZSR:

yzs(t) = 2yo΄(t) +yo(t)= -2e

-t + 12 e

-2t -10e

-3t

To find total Response:

y(t) = yzs(t) + yzi(t)

= -e-t +11e

-2t – 10e

-3t

4) Find the impulse response of the system y΄΄(t) + 3y΄(t) +2y(t) = x΄΄(t).

Solution:

Characteristic equation: s2+3s+2 = 0

Roots are: s1 = -1, s2 = -2

Natural Response:ho(t) =K1e

-t + K2e

-2t;

K2 = -1, K1 = 1


7/36

ho(t) = (e-t -e

-2t)u(t)

The required impulse response is then h(t) = ho΄΄(t)

h(t) = (e-t – 4e

-2t)u(t) + δ(t)

5) Consider the first-order system y΄(t) + 2y(t) = x(t). Find its response if x(t) =

cos(2t), y(0) = 2

Solution:Characteristic equation: s + 2 = 0

Root: s = -2

Natural response: yN(t) = Ke-2t

Forced response:X(t) = cos(2t), So yF(t) = Acos(2t) + Bsin(2t)

A = 0.25, B = 0.25

yF(t) = 0.25cos(2t) + 0.25sin(2t)

Total Response:y (t) = yN(t) + yF(t)

= Ke-2t

+ 0.25cos(2t) + 0.25sin(2t)

With y(0) = 2, K = 1.75

Y(t) = [1.75e-2t

+ 0.25cos(2t) + 0.25sin(2t)]u(t)

Unit – II

1. Find the energy in the signal x[n] = 3(0.5)n, n≥0

E = ∑∞

−∞=n

2 ]n[x

= ∑∞

=0

2)5.0(3n

n ⇒ ( ) J1225.01

925.09

n

0n

⇒−

=∑∞

=

2.

Define folding operation of discrete signals

The signal y[n] = x[-n] represents a folded version of x[n], a missor image

of the signal x[n] about the origin n =0. The signal y[n] = x[-n-α] may be obtained

from x[n] in one of two ways

a. x[n]→ delay (shift right) by α → x[n-α] → fold → x[-n - α]

b. x[n]→ fold → [-n] → advance (shift left) by α → x[-n -α]

3. Let x[n] = {2,3,4,5,6,7} find the following y[n] = x[n-3], f[n] = x[n+2]


8/36

a. y[n] = x[n-3] = {0,2,3,4,5,6,7}

b. f[n] = x[n+2] = {2,3,4,5,6,7}

4. Give the expression for even and odd parts of the signal

even part = xe [n] = 0.5 x[n] = 0.5 x[-n]

odd part = xo [n] = 0.5 x[n] – 0.5 x[-n]

5. Give the steps for decimation and interlocution by a factor N.

Decimation:

Keep every Nth

sample (at n = KN), this leads to potential loss of information.

Interpolation:

Insert (N-1) new values after each sample. This new sample values may equal

zero (zero interpolation), or the previous value (step) or linearly interpolated values.

6. Let x[n] = { x, 4, 5, 6}. Find y[n] = x[24/3] assceming step interpolation where

needed.

Interpolation:

= { }6,6,6,5,5,5,4,4,4,3,3,33 =n x

Decimation:

= ]3

n2[x = {3,3,4,5,5,6}

7. Write the steps for generating fractional delays

Fractional delay of x[n] requires interpolation, shift and decimation in that order

For [ ]

−⇒

N

Mnxnx =

−

N

MNnx

Interpolate x[n] by N, delay by M, then decimate by N.

8. Let x[n] = {2, 4, 6, 8}. Find the signal y[n] = x [n-0.5] assuming linear interpolation

where needed.

x[n-0.50 =

−

2

1nx

= x

−

2

1n2


9/36

2nx = {2, 3, 4, 5, 6, 7, 8, 4} [Lincar Interpolation]

−

2

1nx = {2, 3, 4, 5, 6, 7, 8, 4} [shifting]

−

2

1n2

x = {3, 5, 7, 4} [decimate by 2]

9. Define digital frequency:

The normalized frequency F = f/s is called the digital frequency and has units of

cycles/ smaple.

Where S = 1/ts corresponds to the sampling sate at interval ts.

10. Is x[n] = Cos (2πFn) periodic if F = 0.32?

The signal is periodic only if its digital frequency F = K/N can be expresed as a

ratio of integrers.

Given F = 0.32 =N

K

25

8

100

32==

The period N = 25

11. What is a principal period?

The range – 0.5 ≤ F ≤ 0.5 defines the principal period or principal range.

12. Define sampling theorem?

For a unique comespondance between an analog signal and the version

reconstucted from its samples, the sampling rate must exceed frice the highest signal

frequency fmax.

13. Define Nyquist rate and Nyquist interval.

The critical rate S = 2fmax is called Nyquist rate or Nyquist frequency and

ts =maxf 2

1is called the Nyquist interval.

14. Define aliasing:

The phenomenon, where a reconstructed sinusoid appears at a lower frequency

than the original is called aliasing. It occurs if the analog signal is sampled below the

Nyquist rate.


10/36

15. Define BIBO stability and its requirements.

Bounded – input, bounded – output (BIBO) stability implies that every bounded

input must result in a bounded output.

Requirement for BIBO stability is every root of the characteristic equation must

have magnitude less than unity.

16. Find the whether the system is stable or not y[n] – y[n-1] = x [n]

The Characteristic equation is z – 1 = 0

Z = 1

The Magnitude equals 1, so the system is unstable.

17. A 100–H2 sinusoid x(t) is sampled at 240 H2. Has aliasing occurred? How many

full periods of x(t) are required to obtain one period of the sampled signal.

Given frequency = 100 Hz

Sampling rate = 240 Hz

The sampling rate exceeds 200 Hz, so there is no aliasing. The digital

frequency12

5

240

100F ==

Thus 5 periods of x(t) yeilds 12 samples (one period) of the sampled signal.

18. Define period of discrete signals?

The period of discrete signals is measured as the number of samples per period.

The period N is always an Integer. For combinations N is the LCM of the individual

periods.

19. Give the operations or discrete signals

1.

Time shift

2.

Folding

20. Give the expression for energy and power in discrete signals

1. Energy: [ ]2

n

nxE ∑α

−∞=

=

2. Power: P = [ ]21N

0n

nxN

1∑−

=


11/36


12/36

Unit – III

1. Define Fourier series?

The Fourier series (Fs) describes a periodic signal xp (t) as a Sum (linear

combination), in the right mix, of harmonics (or sinusoid) at the fundamental

frequency of at xp and its multiples kf o

2. What are the three forms of Fourier series

i.i.i.i. Trignometric form:Trignometric form:Trignometric form:Trignometric form:

( ) ( ) ( )tkf 2Sinbtkf 2cosaatx 0k 1k

0k 0p π+π+= ∑∞

=

ii.ii.ii.ii. Polar formPolar formPolar formPolar form

( ) ( )∑∞

=

++=1

02k

k k o p t kf CosaC t x θ π

iii.iii.iii.iii. Exponential form:Exponential form:Exponential form:Exponential form:

( ) ∑∞

∞=

π×=k

kfot2 j

p e]k [tx

3. Give the expressions for trignometric fourier series coefficients

∫= To dt)t(xT1

a

∫ π=

Tk dt)kfot2(Cos)t(x

T

2a

∫ π= Tk dt)kfot2(Sin)t(sT2

b

4. Give the expressions for exponential fourier series coefficients

[ ] ∫= T dt)t(xT1

0X

X[k] = ∫ π−

T

kfot2 jdte)t(s

T

1

5. Give the difference between the F-form and ω-form of the fourier Transform:

• If the transform contains no impulses:

H (f) and H(ω) are related by ω= 2πf

• If it contains Impulses:

Replace δ(f)by 2π δ (ω) land 2πf by ω elsewhere) to get H (ω)


13/36

6. Find the Fourier transform of Unit impulse function:

Unit impulse is δ (t)

x(t) = δ (t)

x(f) =

∫

∞

∞−

π−δ dte)t( ft2 j

r (t) = 1, t = 0

0, t ≠ 0

x(f) = 1

7. What are the there representations of a relaxed LIT system

i. Differential equation

ii. Transfer function

iii. Impulse response

8. What are the there Basle Fourier transform pairs:

i. 1)t( ⇔δ

ii. rect (t) ⇔ Sinc (f)

iii.f 2 j

1e t

π+α⇔α−

9. Define similarity theorem?

If x(t) ⇔ x(f) then x(t) ⇔ x(-f)

10. Find the Fourier transform of a Decaying exponential

Decaying exponential = e-αt

u (t)

x(t) = e-αt

u (t)

x(f) = ∫∞

π−α−

o

ft2t dtee =f 2 j

1

π+α

e-αt u(t) ⇔ f 2 j

1π+α

11. Define Transfer Function:

The transfer function is a frequency-Domain description of a relaxed LTI

systems. It is also defined as H(f)=Y(f)/X(f), (ie)the ratio of the transformed output and

transformed input.


14/36

12. Give the steps to find the steady state response of an LTI system to a sinusoidal

input

i. Input : Given Input is a sinusoidal input

(ie) x (t) = A Cos (ωot + θ)

ii. Transfer function : Evaluate H(ω) at ωo as k

iii. Steady state output : yss (t) = KA Cos (ωot + θ+ φ)

13. Give the steps to find the zero state Response of LTI systems:

i. Input : Evaluate

ii. Transfer unction : Evaluate H (f) from h (t) or spent equation

iii. Zero-state output : Evaluate Y (f) = X (f) H(f) and find its

inverse Fourier transform to find y(t)

14. What are the types of filter

1. Low pass filter (LPF)

2. High pass fitter (HPF)

3. Band pass fitter (BPF)

4.

Band Stop filter (BSF)

15. Define:

1.1.1.1. Frequency Frequency Frequency Frequency- -- -Selective filter:Selective filter:Selective filter:Selective filter:

Device the passes a certain range of frequencies and blocks the rest.

2.2.2.2. Pass band:Pass band:Pass band:Pass band:

Range of frequencies passed defines the passband

3.... Stop band :Stop band :Stop band :Stop band :

Range of frequencies blocked defines the stop band

4.4.4.4. Cut offCut offCut offCut off frequencies frequencies frequencies frequencies::::

Band- edge frequencies are called the cut off frequencies

5.5.5.5. Amplitude distortion: Amplitude distortion: Amplitude distortion: Amplitude distortion:

Gain is not constant over the required frequency range

6.6.6.6. Phase distortion:Phase distortion:Phase distortion:Phase distortion:

Phase shift is not linear with frequency.


15/36

16. Define phase delay and Group delay

1. Phase delay: tp =( )

ω

ωθ−. The delay at a single frequency

2. Group delay: tg =( )

ω

ωθ−

d

d. The delay for a group of frequencies in a

signal

17. Define: Half –power frequency:

The frequency f =c2

1

π is called the half –power frequency because the output

of a sinusoid at this frequency is only half the input power.

18. Define Half-power Bandwidth?

The frequency rangec2

1f 0π

≤≤ defines the half-power bandwidth over which

the magnitude is les than or equal to2

1 times the peak magnitude.

19. What are the basic lap lace transform pairs;

1. 1)t( ⇔δ

2.

s

1)t(u ⇔

3.

2s

1)t(r ⇔

4. ∞+

⇔∞−

s

1)t(ue

t

20. Give the derivative property of laplace transform

)o(x)s(S)t(x' −×=

)o(x)o(Sx)s(S)t(x'2'' −−×=

For zero Ic: )s(S)t(x nn ×⇔


16/36

Unit-1V

1. Define DTFT and give its two forms:

The discrete Time fourier Transform (DTFT) describes the Spectrum of discrete

time single and formalize the concept that discrete time single have periodic spectra

1. F-form:

[ ]∑∞

∞=

π−=k

kF2 j

p eKX)F(X , [ ] ∫ π= 2

1

21

nF2 j

p dFe)F(XnX

2. Ω ΩΩ Ω - Form:

[ ]∑∞

∞=

Ωπ−=Ωk

k 2 j

p eKX)(X , [ ] ∫π

π−

Ωπ ΩΩπ

= de)(X2

1nX

n2 j

p

2. Give the difference between F-form and Ω form of the DTFT

• If the DTFT contains no impulses: H(F)and H(Ω ) are related by

related by Ω = 2πF.

• If the DTFT contains impulse: Replace δ(F) by 2πδ(Ω) (and 2πF by Ω

else where) to get H (Ω).

3. Write the steps to identify a filter:

Traditional filters can be identified by finding the Gain at Dc and F= 0

(or Ω = π)

From transfer function:

Evaluate H(f) at F = 0 and F = 0.5, or evaluate H (Ω) at Ω = 0 and Ω = π.

From impulse response:

Evaluate ∑h[n] = H(o) and

4. What is the DTPT of a Discrete-time periodic signal.

If x p[n] is periodic with period N and its one- period DTFT is x 1[n]⇔ x1[f],

then

( )∑−

=

−δ=⇔1N

0K

opp Fk f N

1)F(X]n[X [ N impulses per period 0 ≤ F


17/36

6. What is the times-n property is DTFT.

Multiply x[n] by n ⇔ Differentiate the DTFT

n x[n] ⇔ dF

)F(dx

2

j p

π (or)

Ω

Ωφ

d

)(x j

p

7. Let Xp(F)⇔ (0.5)n u[n]= x[n]. Find the time signal corresponding to Yp (F)= XP(F) ⊕

XP(F)

By convolution property:

Multiplication in one domain corresponds to convolution in the other.

Given:

Yp(F) = XP (F) ⊗ Xp (F)

y[n] = x[n].x[n]

= (0.5)n u[n]. (0.5)

n u[n]

y [n]=(0.25)n u[n]

8. Define Frequency Response

Plots of the magnitude and phase of the transfer function against frequency are

referred to collectively as the frequency Response. Frequency Response is a very useful

way of describing and identifying digital fitters.

9. Give the steps to find steady-state response to a Discreet-Time Harmonics.

1.1.1.1. Input:Input:Input:Input:

Given input is sinusoidal and if the form

x[n]= A cos (2πn F0 + θ)

2.2.2.2. Transfer function:Transfer function:Transfer function:Transfer function:

Evaluate HP(F) at F=F0 as Ho φ

3.3.3.3. Steady state o/p:Steady state o/p:Steady state o/p:Steady state o/p:

The steady state o/p is given as

yss[n]=A Ho cos (2πn Fo + θ+φ)

10. Design a 3-point FIR fitter with impulse response h[n] {∞, β, α} that completely

blocks the frequency F=3

1 and passes the frequency F=0.125 with unit gain. What is

the dc gain of this filter?

Solution:

Filter transfer function is F2 jp e)F(H πα= + F2 je π−α+β

= β +2 α Cos (2πF)


18/36

Given:

It blocks the frequency F =3

1

H (3

1 ) = 0

1. H

3

1 = )

32(Cos2 πα+β = 0

= β - α = 0

2. H(0.125) = 1

= β + 2α Cos (2π (0.125) = 1

β - α 2 = 1

β = α = 0.414

h[n] = {0.414, 0.414, 0.414}

Dc gain is, H(0) = ∑h[n]

= 0.414 + 0.414 + 0.414

H(0) = 1.2426

11. Give the Times n property of Z transform.

hx[n] ⇔ -Z

dz

)z(dx

12. Define pole zero plot:

A plot of the poles and zeros of a rational x(z) in the Z-plane constitutes a pole-

zero plot and provides a visual picture of the root locations.

13. Find the Z- transform for the sequence x[n]={-7, 3, 1, 4, -8, 5}

x(z) = ∑∞

α=

−

k

k z]k [x

x(z) = -7z2

+ 3z1

+ z0

+ 4z-1

– 8z-2

+ 5z-3

14. Write the steps to solve the difference equations using the Z-transform

1.1.1.1. Relaxed system:Relaxed system:Relaxed system:Relaxed system:

Transform the difference equation, then find Y(z) and its inverse.

2.2.2.2. Not relanced:Not relanced:Not relanced:Not relanced:

Transform using the shift property and initial condition. Find Y(z) and its

inverse.


19/36

15. How can we analyze system using the transfer function is Z-transform?

Zero-stare Response: Evaluate Y(z)=X(z) H(z) and take inverse

transform.

Zero- input Response: Find difference equation. Transform this using

the shift property and initial conditions. Find the response in the Z-

domain and take inverse transform.

16. Write the steps to find the steady state response in z-transform.

1. Input : Given input is of sinusoidal

(ie) x[n]=A cos (2πnF0 + θ)

2. Transfer function : Evaluate H(z) at 0F2 j

eZ π= as Ho

3. Steady state output : The output is given as

y[n]=AH0 Cos (2πnF0 + θ + φ0)

17. Find the Z-transform of unit step function.

Solution:

x[n] = u[n]

x(z) =

∑

∞

=

−

0k

k z

= ∑∞

=

−

0K

K1)Z( =1z1

1−−

=1z

z

−

18. Define Region of Convergence

The value of z for which it does converge defines the region of convergence

(ROC) for X (z).

19. Define:

Scaling property = ( )α

⇔α zX]n[xn

Convolution property = x[n] * h[n] = x[n]. h[n]

Folding property = x[-n] ⇔ x )z

1(


20/36

20. Find the Z-transform of y[n] = nu[n]

Solution:

Using times-n property

Y(z) =

−− 1z

z

dz

dz

= -z( )

−+

−

−

1z

1

)1z(

z2

=( )21z

z

−


21/36

UNIT V

1) Give the expression for N-point DFT and N-point IDFT

XDFT[k] = x[n] e-j2πnk/N

, k = 0, 1, 2,……., N – 1

X[n] = 1/N XDFT[k] e j2πnk/N , n = 0, 1, 1,……., N – 1

2)

Give the steps to generate one period of a circularly shifted periodic signal.

a.

To generate x[ n – no ]: Move the last no samples of x[n] to beginning.

b. To generate x[ n + no ]: Move the first no samples of x[n] to end.

3) Let y [n] = {0, 1, 2, 3, 4, 5, 6, 7}, n = 0, 1, ……, 7. Find one period of the

circularly shifted signals f[n] = y[n – 2]

Solution:To create f[n] = y[n – 2], move the last two samples to the beginning.

So,

y[n – 2] = {6, 7, 0, 1, 2, 3, 4, 5}

4)

Let x[n] = {2, 3, 2, 1} and XDFT(k) = {8, -2j, 0, 2j}. Find the DFT of the 12-

point signal described by y[n] = {x[n], x[n], x[n]}.

Solution:

Signal replication by 3 leads to spectrum zero interpolation and

multiplication by 3. Then,

YDFT(k) = 3XDFT[k/3] = { 24, 0, 0, -6j, 0, 0, 0, 0, 0, j6, 0, 0}.

5)

Define Decimation in DFT.

If we choose the number N of samples as N = 2m, we can reduce the

computation of an N-point DFT to the computation of 1-point DFTs in

m-stages and the 1-point DFT is just the sample value itself. This

process is called decimation.

6)

Define Decimation-in-Frequency algorithm.

The DIF FFT algorithm starts by reducing the single N-point transform

at each successive stage to two N/2-point transforms, then N/4-point

transforms and so on, until we arrive N 1-point transform that

corresponds to the actual DFT.

7) How unequal lengths affect the DFT result?

1) If M=N, the IDFT is periodic with period M, and its one period

equals the N-sample x[n]. Both the DFT matrix and IDFT matrix are

∑

n=0

N-1

∑ k=0

N-1


22/36

square (M X N), and allow a simple inversion relation to go back

and forth between the two.

2) If M > N, the IDFT is periodic with period M. Its one period is the

original N-sample x[n] with M – N padded zeros. The choice M>N

is equivalent to using a zero padded version of x[n] with a total of M

samples and M X N square matrices for both the DFT matrix and

IDFT matrix.

8) Define Decimation-in-Time algorithm.

In DIT FFT algorithm, start with N 1-point transforms combine adjacent

pairs at each successive stage into 2-point transforms, then 4-point

transforms and so on, until we get a single N-point DFT result.

9) Define Vandermonde matrix.

The elements of the DFT and IDFT matrices satisfy Aij = A(i – 1)(j – 1).

Such matrices are known as Vandermonde matrix.

10)

Give the advantage of FFT over DFT.

Fast algoritms reduce the problem of calculating N-point DFT to that of

calculating many smaller size DFTs.

11)

What are the useful DFT pairs.

{1, 0, 0, ….. ,0}(impulse) {1, 1,1 ….. ,1} (constant)

{1, 1, 1, ….. ,1}(constant) (N, 0, 0, …. , 0}( impulse)

αn(exponential) (1 - αn) /(1 – e-j2πk/N)

cos(2πnko/N)(sinusoid) 0.5Nδ[k-k o] + 0.5Nδ[k – (N-k o)] (impulse pair)

12)

How can we compute a N-point periodic convolution y[n] = x[n] h[n].

i.

Compute their N-sample DFTs XDFT[k] and HDFT[k].

ii. Multiply them to obtain YDFT[k] = XDFT[k]HDFT[k].

iii. Find the inverse of YDFT to obtain the periodic convolution y[n].

13) How can we compute regular convolution.

For two sequences of length M and N, the regular convolution contains

M+N-1 samples. We must thus pad each sequence with enough zeros, to

make each sequence of length M+N-1, before finding DFT.

14) Give the method to implement periodic correlation

Periodic correlation can be implemented using the DFT by taking two

conjugation steps prior to taking inverse DFT. The periodic correlation

of two sequences x[n] and h[n] of equal length N gives,

rxh[n] = x[n] h[n] XDFT[k] HDFT*[k]


23/36

15) Give the difference between the DIF and DIT.

a. In DIF, input sequence x[n] occurs in normal form, while output X(k)

appears in bit reversad order. In DIT, input sequence x[n] appears in bit

reversed order while output X(k) appears in normal form.

b.

DIF butterfly is slightly different from DIT, where in DIF complex

multiplication takes place after the add and subtract operation. In DIT,

complex multiplication takes place before add and subtract operation.

c. Number of additions and multiplication are same in both DIT and DIF

algorithm.

16)

Give symmetry property.

WNk + N/2

= - WNk

17) What are the two algorithms in FFT.

a.

Decimation-in-Time (DIT) algorithm.

b.

Decimation-in-Frequency (DIF) algorithm.

18)

Define periodicity property.

WN k + N

= WNk

19) Explain about the speed of fast convolution.

A direct computation of the convolution of two N-sample signals require

N2 complex multiplications. The FFT method works with sequences of

length 2N. the number of complex multiplications involved is

2(Nlog22N) to find the FFT of the two sequences, 2N to form the

product sequence, and Nlog2N to find the IFFT sequence that gives

convolution. It thus requires 3N log2 2N + 2N complex multiplications.

If N = 2m, the FFT approach becomes computationally superior only for

m > 5 or so.

20) Let y [n] = {0, 1, 2, 3, 4, 5, 6, 7}, n = 0, 1, ……, 7. Find one period of the

circularly shifted signals f[n] = y[n + 2]

Solution:

To create f[n] = y[n + 2], move the first two samples to the end. So,

y[n + 2] = {2, 3, 4, 5, 6, 7, 0, 1}


24/36

UNIT II

1) Let x[n] = {2, 3, 4, 5, 6, 7}. Find and sketch the following.

a) y[n] = x[n-3] b) f[n] = x[n+2] c) g[n] = x[ - n]

d) h[n] = x[-n+1] e) s[n] = x[-n-2]

Solution:

a)

y[n] = x[n-3] ={0, 2, 3, 4, 5, 6, 7}

b)

f[n] = x[n+2] = {2, 3, 4, 5, 6, 7}

c) g[n] = x[ - n] = {7, 6, 5, 4, 3, 2}

2

7

x[n+2]

n

-4 1

2

7

x[n-3]

n

6

2

7

x[n]

n

-2 3


25/36

d) h[n] = x[-n+1] = {7, 6, 5, 4, 3, 2}

d) s[n] = x[-n-2] = {7, 6, 5, 4, 3, 2}

2) Consider the recursive digital filter whose realization is shown. What is the

response of this system if x[n] = 6u[n] and y[-1] = -1, y[-2] = 4.

0

7

n

-5

2

4

7

x[-n+1]

n

-2

2

2

7

x[-n]

n

-4

2

x[-n+1]


26/36

Solution:

The differential equation is y[n] – y[n-1] – 2y[n-2] = 2x[n] – x[n-1]

Characteristic equation: z2 – z – 2 = 0

Roots are : z1 = -1 and z2 = 2

Natural Response: yN[n] = A(-1)n + B(2)n

ZIR:

y[-1] = -1, y[-2] = 4

and A= 3, B = 4

yzi[n] = 3(-1)n + 4(2)

n

To find yo[n]:

Forced response : yF[n] = C

C = -3

yF[n] = -3

yo[n] = (-1)n + 8(2)

n – 3 ( using zero initial conditions )

ZSR:

yzs[n] = 2yo[n] – yo[n-1]

= [2(-1)n +16(2)

n – 6]u[n] – [ (-1)

n-1 + 8(2)

n-1 – 3]u[n-1]

Total Response:

Y[n] = [3(-1)n + 4(2)

n]u[n] + [2(-1)

n +16(2)

n – 6]u[n] –[ (-1)

n-1 + 8(2)

n-1 – 3]u[n-

1]

3) Consider the difference equation y[n]-0.5y[n-1] =5cos(0.5nπ), n ≥ 0, withy[ -1] =

4.

Solution:Characteristic equation: z – 0.5 = 0

Roots are : z = 0.5

Natural Response:

yN[n] = K(0.5)n

Z-1

Z-1

∑

∑

∑ X[n] y[n]

+

+

+

+

+

-

2

2


27/36

Since x[n] = 5cos(0.5nπ) , the forced response is yF[n] = A cos(0.5nπ) +Bsin(0.5nπ)

Evaluating the equation:

A = 4, B = 2.

yF[n] = 4 cos(0.5nπ) + 2 sin(0.5nπ)Total response:

y[n] = K(0.5)n + 4 cos(0.5nπ) + 2 sin(0.5nπ)

With y[-1] = 4, K = 3

Thus,

y[n] = 3(0.5)n + 4 cos(0.5nπ) + 2 sin(0.5nπ)

4) Consider the difference equation y[n]-0.6y[n-1] = (0.4)n, n≥0, with y[-1] =

10. Find the total response.

Solution:Characteristic equation: z – 0.6 = 0

Roots are : z = 0.6

Natural Response:

yN[n] = K(0.6)n

Since x[n] = (0.4)n, the forced response is yF[n] =C(0.4)

n

C = -2

yF[n] = -2(0.4)n

Total response:

y[n] = K(0.6)n

- 2(0.4)n

With y[-1] = 10, K = 9

Thus,

y[n] = 9(0.6)n - 2(0.4)

n

5) Find the impulse response of y[n] –1/6 y[n-1] – 1/6 y[n-2] = 2x[n] – 6x[n –

1].

Solution:Characteristic equation: z

2– 1/6z – 1/6 = 0

Roots are : z1 = ½ Z2 = -1/3

Natural Response h[n] = K1(1/2)n + K2(-1/3)

n


28/36

With h[0] = 1 and h[-1] =0,

K1 = 0.6 and K2 = 0.4

To find ho[n]:

ho[n] = 0.6(1/2)n + 0.4(-1/3)

n

Impulse Response:

h[n] = 2ho[n] – 6ho[n-1]

h[n] = [1.2(1/2)n + 0.8(-1/3)

n]u[n] – [3.6(1/2)

n-1 + 2.4 (-1/3)

n-1]u[n-1]

UNIT III

1) Give the properties of Fourier transform.

Property x(t) (t) (t) (t) X( f) f) f) f)

Similarity x(t) X(-f)

Scaling x(αt) 1/(|α|) X(f/ α)

Folding x(-t) X(-f)

Time Shift x(t-α) e-j2πf αX(f)

Frequency Shift e j2παt

x(t) X(f-α)

Convolution x(t)h(t) X(f)H(f)

Multiplication x(t)h(t) X(f)H(f)

Modulation x(t)cos(2παt) 0.5[X(f+α) + X(f-α)]

Derivative x΄(t) j2πfX(f)

Times-t - j2πtx(t) X΄(f)

Correlation x(t)y(t) X(f)Y*(f)


29/36

2) Give the properties of Laplace transform.

Property x(t) X(s)

Superposition αx1(t) + βx2(t) αX1(s) + βX2(s)

Times-exp e-αt

x(t) X(s + α)

Times-cos Cos(αt)x(t) 0.5[X(s + jα) + X(s – jα)]

Times-sin sin(αt)x(t) j0.5[X(s + jα) - X(s – jα)]

Time Scaling x(αt), α > 0 (1/ α)X(s/ α)

Time Shift x(t- α)u(t- α) , α > 0 e-αsX(s)

tx(t) - [dX(s)/ds]Times-t

tnx(t) (-1)

n [d

nX(s)/ds

n]

x΄(t) sX(s) – x(0 - )

x΄΄(t) s2

X(s) – sx(0-) - x΄(0-)Derivativex(n)(t) snX(s) – sn-1x(0-) - …. – xn-1(0-)

Convolution x(t)h(t) X(s)H(s)

3) Explain briefly about frequency response of filters?

Frequency selective filter is a device that passes certain range of

frequencies called pass band and blocks the rest called stopband.

Types:1. Lowpass filter

Impulse response is hLP(t) = 2f c sinc(2f ct)

Transfer function HLP (f) = rect (f/2f c)

2. Highpass filter

Impulse response is hHP (t) = δ (t) - 2fc sinc (2f ct)

Transfer function HHP (f) = 1 - rect (f/2f c)


30/36

3. Bandpass filter

Impulse response is hBP (t) = 4f c sinc(2f ct) cos(2πf ot)

Transfer function HBP (f) = rect(f + f o /2f c) + rect(f - f o /2f c)

4. Bandstop filter

Impulse response is hBP (t) = δ (t) - 4f c sinc(2f ct) cos(2πf ot)Transfer function HBP (f) = 1 - rect (f + f o /2f c) + rect (f - f o /2f c)

4. Let y΄΄(t) + 3y΄(t) + 2y(t) = 4e-2t

, with y(0) = 3 and y΄(0) = 4. Find the total

response y(t) ?

Solution:Transformation to the s-domain using the derivative property yields

s2

Y(s) – s y(0) - y΄(0) + 3 [s Y(s) – y(0)] + 2Y(s) = 4/(s+2)

Substitute for initial conditions and rearranging

Y(s) = (3s2 + 19s + 30) / ((s + 1) (s + 2)2)

Using partial fraction

Y(s) = [K1 / (s + 2)] + [K2 / (s + 2)2] + [K3 / (s + 2)]

Solving for the constants

K1 = 14, K2 = -4, K3 = -11

Taking inverse transformation

y (t) = (14e-t – 4te-2t – 11e-2t) u(t)

5. Consider a system whose transfer function is H(s) = 1/ (s2 + 3s + 2). Find its zero

state, zero input, total response, assuming the input x (t) = 4e-2t and the initial

conditions y(0) = 3 and y΄(t) = 4.

Solution:

ZSR:

Transform x (t) to X(s) = 4/(s + 2) and Yzs(s) = H(s) X(s)

Yzs(s) = 4/ [(s + 2) (s2 + 3s + 2)] = [K1 / (s + 2)] + [K2 / (s + 2)

2] + [K3 / (s + 2)]

Solving for the constants

K1 = 4, K2 = -4, K3 = -4

Taking inverse transformation

yzs(t) = (4e-t – 4te-2t – 4e-2t) u(t)


31/36

ZIR:

(s2 + 3s + 2) Y(s) = X(s), from this obtain the system differential

equation

y΄΄(t) + 3y΄(t) + 2y(t) = x(t)

Assuming zero input and non-zero initial conditions, this transforms to

s2 Yzi(s) – s y(0) - y΄(0) + 3 [s Yzi(s) – y(0)] + 2Yzi(s) = 0

With the given initial conditions

Yzi(s) = (3s + 13)/ (s2 + 3s + 2) = K1 / (s + 1) + K2 / (s + 2)

By partial fraction

K1 = 10, K2 = -7

Taking inverse transforms

Yzi(t) = (10e-t – 7e-2t) u(t)

Total response

y (t) = yzs(t) + yzi(t)

y (t) = (14e-t – 4te

-2t – 11e

-2t) u(t)

UNIT IV

4) Give the properties of DTFT.

Property DT Signal Result (F-Form) Result (Ω-Form)

Folding x[-n] Xp( -F) = Xp*(F) Xp( -Ω) = Xp*( Ω)

Time shift x[n-m] e-j2πmF

Xp(F) e-j Ω m

Xp(Ω)

Frequency shift e j2πnFo

x[n] Xp(F – Fo) Xp(Ω – Ωo)

Half-period shift (-1)nx[n] Xp(F – 0.5) Xp(Ω – 0.5)

Modulation Cos(2πnFo)x[n] 0.5[Xp(F +Fo) + Xp(F – Fo) 0.5[Xp(Ω + Ωo) + Xp(Ω – Ωo)

Convolution x[n]y[n] Xp(F)Yp(F) Xp(Ω ) Yp(Ω)

Product x[n]y[n] Xp(F)Yp(F) Xp(Ω)Yp(F)

Times-n nx[n] (j/2π)[dXp(F)/dF] j[dXp(Ω)/d Ω]

5) Let one period of xp[n] be given by x1[n] = {3, 2, 1, 2}, with N = 4. Find itsDTFT.


32/36

Solution:

Its DTFT is X1(F) = 3 + 2e-j2πF

+ e-j4πF

+ 2e-j6πF

The four samples of X1( kFo ) over 0 ≤ k ≤ 3 are

X1( kFo ) = 3 + 2e-j2πk/4

+ e-j4πk/4

+ 2e-j6πk/4

= {8, 2, 0, 2}

The DTFT of the periodic signal xp[n] for one period 0 ≤ F ≤ 1 is thusXp(F) = ¼ X1( kFo)δ( f- k/4 ) = 2δ(F) + 0.5δ( F – ¼ ) + 0.5δ(F – ¾ )

6) Give the properties of Two-sided Z- Transform.

Property Signal z-Transform

Shifting x[n-N] z–N

X(z)

Reflection x[ -n ] X(1/z)

Anti-causal x[ -n ]u[ -n – 1] X(1/z) – x[0]

Scaling αnx[n] X(z/ α)

Times-n nx[n] - zdX(z)/dz

Times-cos cos(nΩ)x[n] 0.5[X(ze jΩ) + X(ze-jΩ)]

Times-sin sin(nΩ)x[n] j0.5[X(ze jΩ) - X(ze-jΩ)]

Convolution x[n]h[n] X(z)H(z)

7) Solve the differential equation y[n] – 0.5y[n – 1]=2(0.25)nu[n] with y[ - 1]= -

2.

Solution:

Transformation using the right shift property yields,

Y(z) – 0.5{z-1

Y(z) + y[-1]} = 2z / (z-0.25)

Y(z) = [z(z + 0.25)] / [(z – 0.25)(z – 0.5)]

Xp( F)

2 2 2

0.5 0.5

0.5-0.5

∑ k=0

3


33/36

Y(z) / z = (z + 0.25) / [(z – 0.25)(z – 0.5)]

= [A / (z – 0.25)] + [ B /(z – 0.5)]

Using partial fraction method,

A = -2, B = 3

Y(z) = [-2z / (z – 0.25)] + [ 3z /(z – 0.5)]

Taking inverse transforms,

y[n] = [ - 2(0.25)n + 3(0.5)

n]u[n]

8) Consider a system described by y[n] = 0.5y[ n – 1] + x[n]. Find its steady

state response to the sinusoidal input x[n] = 10cos(0.5nπ + 60o) to discrete

time harmonics.

Solution:

The transfer function Hp(F) is given by,

Hp(F) = 1 / [1 – 0.5e-j2πF]

Evaluate Hp(F) at the input frequency F = 0.25

Hp(F) = 1 / (1 + 0.5j) = 0.4 √5 - 26.6o = 0.8944 - 26.6o

The steady-state response then equals

yss(n] = 10(0.4 √5) cos(0.5nπ + 60o- 26.6o )

= 8.9443 cos(0.5nπ + 33.4o )

UNIT V

1) Let x[n] = {1,2,1,0}. With N = 4 find the DFT.

Solution:XDFT[k] = ∑ x[n] e

-j2πnk/N

Successively compute:

K = 0; XDFT[0] = x[n]e0 = 1+2+1+0 = 4

K = 1; XDFT[1] = x[n]e

-jnπ /2

= -j2

K = 2; XDFT[2] = x[n]e-jnπ

= 0

K = 3; XDFT[3] = x[n]e-j3nπ /2

= j2

The DFT is thus XDFT[k] = {4, -j2, 0, j2}

2) Given x[n] = {0, 1, 2, 3, 4, 5, 6, 7} Find X(k) using DIT FFT algorithm.

n=0

N-1

∑ n=0

3

∑ n=0

3

∑ n=0

3

∑ n=0

3


34/36

Solution:

W80 = 1, W8

1 = 0.707 - 0.707j, W8

2 = -j, W8

3 = -0.707 -

0.707j

X(k) = {28, (-4+9.656j), (-4+4j), (-4+1.656j), -4, (-4-1.656j), (-4-4j), (-4-9.656j)}

3) Given x[n] = n+1, N=8. Find X(k) using DIF FFT algorithm.

Solution:

W80 = 1, W8

1 = 0.707 - 0.707j, W8

2 = -j, W8

3 = -0.707 -

0.707j

X(k) = {36, (-4+9.656j), (-4+4j), (-4+1.656j), -4, (-4-1.656j, (-4-4j), (-4-9.656j)}

x(0)=0

x(4)=4

x(2)=2

x(6)=6

x(1)=1

x(5)=5

x(3)=3

x(7)=7

W80

W80

W80

W80

W80

W82

W82

W80

W80

W81

W82

W83

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

W80

W81

W82

W80

W82

W80

W80

W80

W80

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

x(0)=1

x(1)=2

x(2)=3

x(3)=4

x(4)=5

x(5)=6

x(6)=7

x(7)=8

X(5)= - 4 - 1.656j

X(3)= - 4 + 1.656j

X(1)= - 4 - 9.656j

28

-4+9.656j

-4+4j

-4+1.656j

-4

-4-1.656j

-4-4j

-4-9.656j

W83 W8

2 W8

0

X(0)=36

X(4)= - 4

X(2)= - 4+4j

X(6)= - 4 - 4j

X(1)= - 4 +9.656j


35/36

4) Given X(k) ={20, (-5.828-2.414j), 0, (-0.172-0.414j), 0, (-0.172+0.414j), 0,

(-5.828+2.414j)}. Find x[n] using DIF FFT algorithm.

Solution:

W80 = 1, W8

-1 = 0.707 + 0.707j, W8

-2 = j, W8

-3 = -0.707+

0.707j

x[n] = {1, 2, 3, 4, 4, 3, 2, 1}

5) Find the IDFT of X(k) = {4, (1-2.414j), 0, (1-0.414j), 0, (1+0.414j), 0,

(1+0.414j)} using DIT FFT algorithm.

Solution:

W80 = 1, W8

-1 = 0.707 + 0.707j, W8-2 = j, W8

-3 = -0.707+

0.707j

X(0)

X(4)

X(2)

X(6)

X(1)

X(5)

X(3)

X(7)

W80

W80

W80

W80

W80

W8-2

W8-2

W80

W80

W8-1

W8-2

W8-3

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

W80

W8-1

W8-2

W80

W8-2

W80

W80

W80

W80

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

X (0)

X(1)

X(2)

X(3)

X(4)

X(5)

X(6)

X(7)

⅛ ⅛

⅛

⅛

⅛ ⅛

⅛

⅛

x

(0)=1

x

(4)=4

x

(2)=3

x

(6)=2

x

(1)=2

x

(5)=3

x

(3)=4x

(7)=1W8

-3 W8

-2 W8

0

⅛ x(0)=1

⅛ x(1)=1

⅛ x(2)=1

⅛ x(3)=1

⅛ x(4)=0

⅛ x(5)=0

⅛ x(6)=0

⅛ x(7)=0


36/36

x[n] = {1, 1, 1, 1, 0, 0, 0, 0}

Documents

Good Question DSP