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) ?

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

(a) 0

(b) 6

(c) 0

(d) -6

- Find the signal power of x(t)

(a) 0.25 W

(b) 0.75 W

(c) 0.5 W

(d) 1 W

- For the above system

G_{1}(s) = s^{-1}/1 + 2s^{-1} + s^{-2}

Find G_{2}(s) for the given system to be invertible

(a) s/1 + 2s + s^{2}

(b) 1 + 2s + s^{2}

(c) 1/1 + 2s + s^{2}

(d) 1 + 2s + s^{2}

- The transfer function of an LTI system may be expressed as

H(z) = k (z – z_{1})……(z – z_{m})/(z – p_{1})………(z – p_{n})

Following are the statements made

- poles of H(z) are called natural modes.
- 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

- If y(t) e
^{x(t)}, then the relation is

(a) dynamic

(b) static

(c) memory

(d) none of these

**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)

- 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

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

(a) E_{1}/2

(b) E_{1}

(c) 2E_{1}

(d) E_{1}/4

- 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 }– e^{2t}] u(t)

(c) sin 2 t

(d) (1 – 4e^{-4t}) u(t)

- 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

- 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

- 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}

- 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)

- 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)

- Consider the transform pair given below

x(t) u(t) – 2s/s^{2} + 2

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

(a) 2se^{-2s}/s^{2} + 2

(b) 2se^{2s}/s^{2} + 2

(c) 2(s – 2)/(s – 2)^{2} + 1

(d) 2(s + 2)/(s + 2)^{2} + 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

- Consider a signal f(t) = 3t
^{2}+ 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

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

(a) time scaling

(b) time shift

(c) time reversal

(d) time expansion

- 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

- For the signal shown below, the region of convergence will be

(a) 0_{1} < 0 < 0_{2} is s-plane

(b) entire s-plane

(c) imaginary axis

(d) entire s-plane except imaginary axis

- 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)}

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

**Codes**

** A B**

**(a) 2 4**

**(b) 4 3**

**(c) 1 3**

**(d) 1 2**

- 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

- A periodic signal x(t) of period t
_{0}is given by

x(t) = {1, |t| < t_{1} 0, t_{1} < |t| < t_{o}/2

The DC component of x(t) is

(a) t_{1}/t_{o}

(b) t_{1}/2t_{o}

(c) 2t_{1}/t_{o}

(d) t_{o}/t_{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

- 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

- The fourier series representation of an impulse train denoted by s(t) = d(t – nt
_{o}) is given by

(a) 1/t_{0} exp – j2 nt/t_{o}

(b) 1/t_{o } exp – j nt/t_{0}

(c) 1/t_{0} exp jnt/t_{o}

(d) 1/t_{0} exp j2nt/t_{o}

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

V_{C} (s) = s + 1/s^{3} + s^{2} + s + 1 = 1/s^{2} + 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

- 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)

- If f(s) = [f(t)] = k/(s + 1) (s
^{2}+ 4) then lim f(t) is given by

(a) k/4

(b) zero

(c) infinite

(d) undefined

- If l[f(t)] = 2(s + 1)/s
^{2}+ 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

- 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}

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

(a) e^{st} f(s)

(b) e^{-st} f(s)

(c) f(s)/1 + e^{st}

(d) f(s)/1 – e^{-st}

- 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

- 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

- X
_{1}(z) = 2z + 1 + z^{-1}and x_{2 }(z) = z + 1 + 2z^{-1}is…………….pair.

(a) even signal

(b) odd signal

(c) time -power signal

(d) none of these

- A linear discrete-time system has the characteristic equation z
^{3}– 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

**Answers with Solutions**

**unit Exercise – 1**

- (b)

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

= 1[-1 + 2] + 2[1 + 1] + 1 [2 – 1]

= 1 + 4 + 1 = 6

- (c)

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

p = 1/t x^{2} (t) = 1/2 1. dt = 0.5 W

- (b)

G_{2}(s) should be inverse of G_{1} (s) then only input can be recovered from the circuit and system will become invertible.

G_{2}(s) = G_{1}^{-1} (s)

G_{1} (s) = s^{-1}/1 + 2s^{-1} + s^{-2}

G_{2}(s) = G_{1}^{-1} (s)

= 1 + 2s + s^{2}/s

- (c)

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

- (b)

y(t) = e^{x(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.

- (a)
- (a)

H(z) = 2-5/2 z^{-1}/(1 – 1/2 z^{-1}) (1 – 2z^{-1})

= 2z^{2} – 5/2 z/z^{2} – 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.

- (d)

we know,

E = X^{2}(t) dt

(E_{1}/2) = X^{2}(t) dt x – x

E_{2} = X^{2} (2t) dt

substituting 2t = T,

2dt = dt

E_{2} = x^{2}(t) dt/2

E_{2} = 1/2 x^{2}(t) dt

but from eq. (i) 1/2 x^{2} (t) dt = E_{1}/2

E_{2} = E_{1}/2 (1/2) = E_{1}/4

- (a)

h_{s}(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)

- (b)

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

output = 1/2 x 2 x 1/2 = 1/2

- (c)

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

- (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)

- (a)

x(z) = z^{2} – 3z/z^{2} + 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] – 1/2^{n }u[n]

- (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 4^{n}] u[n]

- (a)

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

- (a)

signal power = 0.5 (2^{2} + 4^{2} + 6^{2} + 2^{2})

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

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

- (c)

f(t) (t – t_{o}) = f(t_{o})

f(t) (t – 2) = f(2)

f(2) = 3(2)^{2} + 2(2) + 1 = 12 + 4 + 1 = 17

- (c)

x_{2} (z) = 4 + 3z + 2z^{2} + z^{3} + z^{4}/1 + 2z + 3z^{2} + 4z^{3} + z^{4}

x_{2}(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}

= 4z^{4} + 3z^{3} + 2z^{2} + z + 1/z^{4} + 2z^{3} + 3z^{2} + 4z + 1

= original z-transform

hence, time reversal property is used.

- (b)

only sine functions, as it has the odd symmetry.

- (b)

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

- (c)
- (c)

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

- (c)

the periodic signal x(t) of period t_{o}

x(t) = 1|t| < t_{1}

x(t) = 0t_{1} < |t| < t_{0}/2

A_{o} = 1/t_{o} x(t) dt = 2t_{1}/t_{o}

- (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.

- (b)

example – A sin 0_{0}t = x(t)

(real and odd function)

x(0) = Aj [(0 + 0_{0}) – (0 – 0_{0})]

(imaginary and odd function of 0)

- (d)

T_{O} (t) = ke^{jk ot}

C_{k} = 1/T_{O} (t) e^{-jk t} dt = 1/t_{o}

t_{o}(t) = 1/t_{o} e^{+ jk t}

- (c)

C = 1/2 F

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

V_{C} (s) = s + 1/s^{3} + s^{2} + s + 1 = s^{2} + 1

I(s) = V_{C}(s)/Z(s) = 1/s^{2} + 1/2 /s

I(s) = s/2(s^{2} + 1)

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

= lim sI (s) = lim s – s/2(s^{2} + 1)

= lim s^{2}/2s^{2} (1 + 1/s^{2})

= 1/2 lim 1/1 + 1/s^{2} = 1/2 A

- (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

- (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, s^{2} + 4 = 0 or s = j2 has polse on j-axis.

- (b)

2(s + 1)/s^{2} + 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} + 0^{2}

where, a = 1, 0 = 2

f(t) = 2e^{-t} cos 2t

f(0) = f(0^{+}) = 2e^{0} cos 0 = 2

f(0) = 2e^{-00} cos 0 = 0

- (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

L^{1}F (s) = 2e^{-t} – e^{-3t}

- (b)

f(t – T) = F(s) e^{-st}

(time shifting property of laplace transform)

- (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

- (c)

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

(N + 1) ROC

- (d)

X_{1} (z) = 2z + 1 + z^{-1}, X_{2} (z) = z + 1 + 2z^{-1}

x_{1 }[k] = {2,1,1}, x_{2}[k] = {1,1,2}

x_{1 }[k] and x_{2}[k] are time reversed pair (mirror image).

- (a)

Linear discrete time system has characteristic equation

z^{3} – 0.81z = 0

z(z^{2} – 0.9^{2}) = 0

z(z – 0.9) (z + 0.9) = 0