# Compute the convolution y[n]* = x[n] * h[n]. when x[n] = (1,2,4), h(n) = {1,1,1,1,1) | compute the convolution y(n)=x(n)*h(n) of the following signals

compute the convolution y(n)=x(n)*h(n) of the following signals , Compute the convolution y[n]* = x[n] * h[n]. when x[n] = (1,2,4), h(n) = {1,1,1,1,1) ?

1. Power spectral density of signal x(t) is shown below. the average power is

(a) 0

(b) 6

(c) 0

(d) -6

1. Find the signal power of x(t)

(a) 0.25 W

(b) 0.75 W

(c) 0.5 W

(d) 1 W

1. For the above system

G1(s) = s-1/1 + 2s-1 + s-2

Find G2(s) for the given system to be invertible

(a) s/1 + 2s + s2

(b) 1 + 2s + s2

(c) 1/1 + 2s + s2

(d) 1 + 2s + s2

1. The transfer function of an LTI system may be expressed as

H(z) = k (z – z1)……(z – zm)/(z – p1)………(z – pn)

1. poles of H(z) are called natural modes.
2. Poles of H(z) are called natural frequencies.

Choose the correct option.

(a) 1 – true, 2 – false

(b) 1 – false, 2-true

(c) 1-true, 2-true

(d) 1-false, 2-false

1. If y(t) ex(t), then the relation is

(a) dynamic

(b) static

(c) memory

(d) none of these

1. Compute the convolution y[n]* = x[n] * h[n]. when x[n] = (1,2,4), h(n) = {1,1,1,1,1)

(a) {1, 3, 7,7,7,6,)

(b) {1,3,3,7,7,6,4}

(c) {1,2,4}

(d) {1,3,7)

1. H(z) = 1/1 – 1/2 z-1 + 1/1 – 2z-1, |z|> 2 is

(a) causal

(b) non-causal

(c) anti-causal

(d) cannot be determined

1. Consider a signal x(t) with energy (E1/2) then time scaling by a factor 2, (i.e., doubling ) will change energy to

(a) E1/2

(b) E1

(c) 2E1

(d) E1/4

1. The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2t u(t). the response of this circuit network to a unit step function will be

(a) 2[1 – e-2t] u(t)

(b) 4[e-t – e2t] u(t)

(c) sin 2 t

(d) (1 – 4e-4t) u(t)

1. The impulse response and the excitation of a linear time-invariant causal system are shown in fig. (a) and (b), respectively. the output of the system at t = 2 s is equal to

(a) 0

(b) 1/2

(c) 3/2

(d) 2

1. An excitation is applied to a system at t = t and its response is zero for – 0 <1 < t . such a system is

(a) non-causal system

(b) stable system

(c) causal system

(d) unstable system

1. The unit impulse response of a linear time-invariant system is the unit step function u(t). for t > 0, the response of the system to an excitation e-at u(t), a > 0 will be

(a) ae-at

(b) (1/a)(1 – e-at)

(c) a (1 – e-at)

(d) 1 – e-at

1. The time signal corresponding to e-2s d/ds 1/(s + 1)2 is

(a) -te-t u(1 – t)

(b) -te-t u(t – 1)

(c) -(t – 2)2 e-(t-2) u(t – 2)

(d) te-t u(t – 1)

1. Consider the tranform pair given below

cos 2t u(t) – x(s).

The time signal corresponding to (s + 1) x(x) is

(a)( cos 2t – 2 sin 2t) u(t)

(b) (cos 2t + sin 2 t/2) u(t)

(c) (cos 2 t + 2 sin 2 t) u(t)

(d) (cos 2t – sin2t/2) u(t)

1. Consider the transform pair given below

x(t) u(t) – 2s/s2 + 2

Determine the laplace transform y(s) of the given time signal in question and choose correct option.

(a) 2se-2s/s2 + 2

(b) 2se2s/s2 + 2

(c) 2(s – 2)/(s – 2)2 + 1

(d) 2(s + 2)/(s + 2)2 + 1

1. Consider a signal x(t) = 2 sin 2t + 4 sin 4t + 6 cos 4t + 2 cos 2t with period 1/2.

the signal power is

(a) 30 W

(b) 36 W

(c) 60 W

(d) 12 W

1. Consider a signal f(t) = 3t2 + 2t + 1 which is multiplied by 2 unit delayed version of impulse and integrated over period – 0 to 0, the resultant is given by

(a) 1

(b) 6

(c) 17

(d) 16

1. Z-transform of the signal is given as 4z4 + 3z3 + 2z2 + z + 1/z4 + 2z3 + 3z2 + 4z + 1 after applying a particular property the z-transform was changed to 4 + 3z + 2z2 z3 + z4/1 + 2z + 3z2 + 4z3 + z4 . the property used is

(a) time scaling

(b) time shift

(c) time reversal

(d) time expansion

1. A periodic triangular wave is shown in the figure . its fourier components will consist only of

(a) neither odd nor even

(b) an odd function

(c) an even function

(d) both odd and even

1. For the signal shown below, the region of convergence will be

(a) 01 < 0 < 02 is s-plane

(b) entire s-plane

(c) imaginary axis

(d) entire s-plane except imaginary axis

1. Match the following list 1 with list 2 and select the correct answer using the codes given below the lists

List 1                                                                                    List 2

A . the fourier transform of g(t – 2) is             1.  G(f) e-j(4f)

1. The fourier transform of g(t/2) is 2. G(2f)
2. 2G(2f)
3. G(f – 2)

Codes

A                             B

(a)               2                             4

(b)              4                              3

(c)              1                              3

(d)             1                               2

1. The trigonometric fourier series of a periodic time function can have only

(a) cosine terms

(b) sine terms

(c) cosine and sine terms

(d) DC and cosine terms

1. A periodic signal x(t) of period t0 is given by

x(t) = {1,  |t| < t1   0, t1 < |t| < to/2

The DC component of x(t) is

(a) t1/to

(b) t1/2to

(c) 2t1/to

(d) to/t1

1. The fourier transform of a function x(t) is x(f). the fourier transform of dx(f)/dt will be

(a) dx(f)/df

(b) j2 fx(f)

(c) jfx(f)

(d) x(f)/jf

1. A signal x(t) has a fourier transform x(0). if x(t) is a real and odd function of t, then x(0) is

(a) a real and even function of 0

(b) an imaginary and odd function of 0

(c) an imaginary and even function of 0

(d) a real and odd function of 0

1. The fourier series representation of an impulse train denoted by s(t) = d(t – nto) is given by

(a) 1/t0  exp – j2 nt/to

(b) 1/to  exp – j nt/t0

(c) 1/t0  exp jnt/to

(d) 1/t0  exp j2nt/to

1. If the laplace transform of the voltage across a capacitor of value 1/2 F is

VC (s) = s + 1/s3 + s2 + s + 1 = 1/s2 + 1

The value of the current through the capacitor at t = 0+ is

(a) zero

(b) 2 A

(c) 1/2 A

(d) 1 A

1. The voltage across an impedence in a network is V(s) = Z(s) I(s), where V(s), Z(s), I(s) are the laplace transform of the corresponding time functions v(t), z(t) and i(t). the voltage v(t) is

(a) V(t) = z(t) V(t)

(b) V(t) = i . z(t -) d

(c) v(t)= i . z(t-) dt

(d)  v(t) = z(t) + i(t)

1. If f(s) = [f(t)] = k/(s + 1) (s2 + 4) then lim f(t) is given by

(a) k/4

(b) zero

(c) infinite

(d) undefined

1. If l[f(t)] = 2(s + 1)/s2 + 2s + 5, then f(0+) and f(0) are given by

(a) 0, 2 respectively

(b) 2, 0 respectively

(c) 0, 1 respectively

(d) 2/5, 0 respectively

1. The inverse laplace transform of the function s + 5/(s + 1) (s + 3) is

(a) 2e-t – e-3t

(b) 2e-t – 2e-3t

(c) e-t – 2e-3t

(d) e-t + e-3t

1. If [f(t)] = f(s), then [f(t – t)] is equal to

(a) est f(s)

(b) e-st f(s)

(c) f(s)/1 + est

(d) f(s)/1 – e-st

1. Lets take 2 signals (-3)k u[k] and u[k – 1] has superimposed. so, the superimposed signal z-transform output is

(a) z/z – 3 + 1/z + 1

(b) z/z + 3 – 1/z – 1

(c) z/z + 3 + 1/z – 1

(d) z/z – 3 – 1/z + 1

1. A system has N different poles. then the system can have

(a) N ROC’s

(b) (N -1) ROC’s

(c) (N + 1) ROC’s

(d) none of these

1. X1 (z) = 2z + 1 + z-1 and x2 (z) = z + 1 + 2z-1 is…………….pair.

(a) even signal

(b) odd signal

(c) time -power signal

(d) none of these

1. A linear discrete-time system has the characteristic equation z3 – 0.8 1z = 0. the system

(a) is stable

(b) is marginally stable

(c) is unstable

(d) stability cannot be assessed from the given information

unit Exercise – 1

1. (b)

p = s(f) d(f) = 1 df + 2df + 1df

= 1[-1 + 2] + 2[1 + 1] + 1 [2 – 1]

= 1 + 4 + 1 = 6

1. (c)

The signal power in x(t) using parseval’s relation is

p = 1/t  x2 (t) = 1/2  1. dt = 0.5 W

1. (b)

G2(s) should be inverse of G1 (s) then only input can be recovered from the circuit and system will become invertible.

G2(s) = G1-1 (s)

G1 (s) = s-1/1 + 2s-1 + s-2

G2(s) = G1-1 (s)

= 1 + 2s + s2/s

1. (c)

The poles of H(z) are called natural modes or natural frequencies.

1. (b)

y(t) = ex(t)

the system represented by the above is static (memoryless) since the output at time t is dependent on t only. further input-output relation is not integrodifferential relation.

1. (a)
2. (a)

H(z) = 2-5/2 z-1/(1 – 1/2 z-1) (1 – 2z-1)

= 2z2 – 5/2 z/z2 – 5/2 z + 1 = 1/2/z – 0.5 + 2/z – 2

h(n) = [(1/2)n + 2n] u(n)

since, h(n) = 0 for n < 0, so it is causal.

1. (d)

we know,

E = X2(t) dt

(E1/2) = X2(t) dt  x – x

E2 = X2 (2t) dt

substituting   2t = T,

2dt = dt

E2 = x2(t)  dt/2

E2 = 1/2 x2(t)  dt

but from eq. (i)  1/2  x2 (t) dt  = E1/2

E2 = E1/2 (1/2) = E1/4

1. (a)

hs(t) = 4e-2t u(t), H(s) = 4/s + 2

for unit step, y(s) = H(s) X(s)

y(s) = 4/s(s + 2) = 2/s – 2/s + 2

y(t) = 2 [1 – e-2t] u(t)

1. (b)

y(t) = x h(t-) d

output  = 1/2 x 2 x 1/2 = 1/2

1. (c)

in the given system, the time t depenes on present value. hence, the system is causal system.

1. (b)

h(t) = u(t), x(t) = e-at u(t)

y(s) = x(s) h(s)

= 1/s + a – 1/s = 1/a [1/s – 1/s + a]

y(t) = 1/a (1 – e-at) u(t)

1. (a)

x(z) = z2 – 3z/z2 + 3/2 z-1 = 1 – 3z-1/1 + 3/2 z-1 – z-2

= 2/1 + 2z-1 – 1/1 – 1/2 z-1, ROC : 1/2 < |z| < 2

x[n]  = 2(2)n u[-n – 1] – 1/2n u[n]

1. (a)

x[n] is right sided.

x(z) = z-1/4z-1/1 – 16z-1

= 49/32/1 + 4z-1 + 47/32/1 – 4z-1

x[n] = [49/32 (-4)n + 47/32 4n] u[n]

1. (a)

x[n] = [n + 6] + [n + 2] + 3 [n] + 2 [n – 3] + [n – 4]

1. (a)

signal power = 0.5 (22 + 42 + 62 + 22)

= 1/2 (4 + 16 + 36 + 4)

= 1/2 (20 + 40) = 30 W

1. (c)

f(t) (t – to) = f(to)

f(t) (t – 2) = f(2)

f(2) = 3(2)2 + 2(2) + 1 = 12 + 4 + 1 = 17

1. (c)

x2 (z) = 4 + 3z + 2z2 + z3 + z4/1 + 2z + 3z2 + 4z3 + z4

x2(1/z) = 4 + 3 (1/z) + 2 (1/z)2 + (1/z)3 + (1/z)4/1 + 2 (1/2) + 3 (1/z)2 + 4(1/z)3 + (1/z)4

= 4z4 + 3z3 + 2z2 + z + 1/z4 + 2z3 + 3z2 + 4z + 1

= original z-transform

hence, time  reversal property is used.

1. (b)

only sine functions, as it has the odd symmetry.

1. (b)

if x(t) is of finite duration and is absolutely integrable, then the ROC is entire s-plane.

1. (c)
2. (c)

the trigonometric fourier series of periodic time function can have sine and cosine terms.

1. (c)

the periodic signal x(t) of period to

x(t) = 1|t| < t1

x(t) = 0t1 < |t| < t0/2

Ao = 1/to  x(t) dt = 2t1/to

1. (b)

fourier transform of x(t) = x(f)

dx(t)/dt = j x(f) = 2 j fx (f)

it is a property of fourier transform.

1. (b)

example – A sin 00t = x(t)

(real and odd function)

x(0) = Aj [(0 + 00) – (0 – 00)]

(imaginary and odd function of 0)

1. (d)

TO (t) = kejk ot

Ck = 1/TO  (t) e-jk t dt = 1/to

to(t) = 1/to  e+ jk t

1. (c)

C = 1/2 F

Z(s) = 1/CS = 2/S

VC (s) = s + 1/s3 + s2 + s + 1 = s2 + 1

I(s) = VC(s)/Z(s) = 1/s2 + 1/2 /s

I(s) = s/2(s2 + 1)

The value of current thorugh the capacitor at t = 0+ is

= lim sI (s) = lim s – s/2(s2 + 1)

= lim s2/2s2 (1 + 1/s2)

= 1/2 lim 1/1 + 1/s2 = 1/2 A

1. (b)

Product in s domain is nothing but convolution in time domain. hence, V(s) = Z(s) I(s) becomes

V(t) =  1 z (t-) d

1. (c)

From final value theorem lim f(t) = lim sF (s) provided s – F(s) has polse with negative real parts, i.e., F(s) is a stable function exponentially decaying to 0 as s – 0

here, s2 + 4 = 0 or s = j2 has polse on j-axis.

1. (b)

2(s + 1)/s2 + 2s + 5 = 2(s + 1)/(s + 1)2 + 4 = 2 [s + 1/(s + 1)2 + 4

this is standard laplace transform for the function.

L|exp (-at) cos t|= s + a/(s + a)2 + 02

where, a = 1, 0 = 2

f(t) = 2e-t cos 2t

f(0) = f(0+) = 2e0 cos 0 = 2

f(0) = 2e-00 cos 0 = 0

1. (a)

f(s) = s + 5/(s + 1) (s + 3) = A/s + 1 + B/s + 3

A = s + 5/s + 3|s = – 1 = 2,  B = s + 5/s + 1|s = – 3

= +2/2 = – 1

L1F (s) = 2e-t – e-3t

1. (b)

f(t – T) = F(s) e-st

(time shifting property of laplace transform)

1. (c)

because superposition means addition of these 2 signals

So, superimposed f[(k)] = (-3)k u[k] + u [ k – 1]

z[(k)] = z/z + 3 + 1/z – 1

1. (c)

For 2 polse, we have 3 ROC conditions, i.e.,

(N + 1) ROC

1. (d)

X1 (z) = 2z + 1 + z-1, X2 (z) = z + 1 + 2z-1

x1 [k] = {2,1,1}, x2[k] = {1,1,2}

x1 [k] and x2[k] are time reversed pair (mirror image).

1. (a)

Linear discrete time system has characteristic equation

z3 – 0.81z = 0

z(z2 – 0.92) = 0

z(z – 0.9) (z + 0.9) = 0