Question

(a) The continuous-time signal x(t) with FT as depicted in the figure shown below is sampled. Sketch the FT of the sampled signal for the following sampling intervals: identify whether aliasing occurs, Ts = 1/12 X(jw) -117 107 W -10 0 117 97 97T (b) Determine the z-transform and ROC for the following time signals: x[n] = (4)"u[n] + (1)"u[ -– 1] Sketch the ROC, poles, and zeros in the z-plane.

Answer 1

Question (a)

The input signal is band limited at

111 rad/s

So the maximum frequency present in the signal is

wm 1171

So

21fm 1170

fm 11 HZ 2

So to avoid aliasing the minimum sampling frequency in Hz should be

fore = 2 x fm sred 71

11 fsread = 2 x 2

fore = 11 Hz s.read

So the sampling frequency should be atleast 11 Hz

The sampling period used is

TE 1 12 S

So the sampling frequency in Hz used is

1 fs 09 T S

f = 12 Hz

Since this is greater than the minimum required sampling frequency of 11 Hz, there wont be aliasing

The digital frequency is

ω Ω fs

So will become

w 9п

№1 9п 3п п, = = = f. 12 4

So will become

wz 1070

1071 577 122 2 fs 12 6

So will become

W3 = 117

11T 23 03 - 12 Fi

So the sketch of the FT of the sampled signal will be

X(el) 12 517 6 31! 11[ 4 12 Зп -[ 111[ 12 70 0 - 4 -5a -5m 6 -12

Question (b)

x[n] (6) a[n] +4). u[-n - 1]

The Z transform is given as

Χ(2) = α[n]z-1 n=-00

-2 Χ(2) = Σ (4) a[n] +() () -η - 1) -- n=-00

σα -2 Χ(2) = Σβ)". u[n]z-1 + Σ4)". u[-η - 1]z- n=-00 n=-00

X(z) = x (2) + X (2)

X,z) = Με u[n]z-n 2. n=- 00

-72 Χ.(z) = Σ(4)". u[-η - 1]z- η =-00

Let evaluate

X,(z)

X,z) = Με u[n]z-n 2. n=- 00

Using

u] = {%, n>0 n<0

We get

O=u Z (+)3 = (2) x -72 u

0=u (-93 =(z)'x -Z 72

Expanding we get

X(z) = 1 +=2-1 + + :-) 2 2

This is an infinite GP with common ratio . The sum of an infinite GP is given as

-1 N r 2

S= 1+r+p2 +p3 +

1 S= 1-r

This is valid only if

Ir] <1

So

X(z) = 1 +=2-1 + + :-) 2 2

1 ROC: <1 X() = 1-12-1 N

1 ROC: X() = 1 -2-1 < N

1 ROC: > X() = 1 -2-1 2 N

n 1 X2) = ROC:[2] > 1 2 2

Now lets evaluate

X (2)

-72 Χ.(z) = Σ(4)". u[-η - 1]z- η =-00

Using

u[-n- 1] = {0; -n-10 -n-1<0

u[-n- 1] = {d} -n > 1 -N <1

u[-n- 1] = {d} n<-1 n> -1

We get

Υ X2(2) = Σ4) - η =-00

2 X2(2) = Σ4--1) η =-00

Expanding we get

(2) = (-) + (-) + (-) + (-) + ..

-3 * (2) = (-) + (:-) + (:-) + (:-) + .......

X (2) = 4z[1 + 4z + (42)2 + (42)3+.

The series in the bracket is an infinite GP with common ratio . The sum of an infinite GP is given as

T= 42

S= 1+r+p2 +p3 +

1 S= 1-r

This is valid only if

Ir] <1

So

X (2) = 4z[1 + 4z + (42)2 + (42)3+.

42 X_(z) = 1- 42 ROC: |4z| <1

1 42 X_(z) = 1-42 ROC: 121 4

So the Z transform is

X(z) = x (2) + X (2)

N 42 1 1 X(2) = + ROC:21 > and 121 1 - 42 2 4 N 2

So we see that the ROC is not possible. So the Z transform does not exist