Question

- Frequency Response (Amplitude Response only). Hz). with frequency, 22. for a discrete time system shown below. *(-1) - x[-2] - ... -0 and yf-1) - Y[-2] ... - x[r] - int) Find “Math Model" for the system. nt) Find "Transfer Function" for the system. Draw the pole-zero plot for the system (use unit circle on Re-Im axis) Sketch the amplitude response of the system → indicate values at important points (92 = 0, 1/4, 21/4, 37/4, T) include detailed steps being used to find values at these points dd analog frequency axis (f = 0/2) below discrete frequency 12, in your esult (x-axis from (b)), when sampling frequency is fs = 1 [MHz]. nalyze the system. Explain your resulting amplitude response in your ow prds → explain cut-off frequency (in analog frequency, J) → explain the roles of poles and zeros

Answer 1

Question (a)

From the diagram, we can write

Y[n] = x[n] + 5yın – 1] - *y[n – 2]

So the difference equation relating the input and the output is

Y[n] = x[n] + 5yın – 1] - *y[n – 2]

Taking Z transform on both sides, we get

(2)47-27 - (2)41-25+ (2)X = (2),

We have used the shifting property of Z transform here, which states that

If

(2)X = {[u]x}

Then

(2)x - 2 = {[êu – 4]}z

(2)47-27 - (2)41-25+ (2)X = (2),

(༤)x = (2): +(2)A- – (2)

(2)x=(:--%++-- 1) (2)

So, the transfer function of the system is

H(z) = 10

z-+*+1-25-I -= (2)

H(2) = GIN Ny +

(€ -2)(- 2) = (2)

Poles are at

o (--)(--)

Zeros are at

0 = 2 0 = 2

So the pole zero plot is

Pole-Zero Map Imaginary Axis -1.5 1.5 5 -0. 0 Real Axis 0.5 1

Question (b)

We have derived the transfer function of the filter in part (a) which is

z-+*+1-25-I -= (2)

To obtain the frequency response, substitute

zrejn

We get

Aram + ܘܙܨܢ -(Holm

e-j@ = cos(O) - j.sin(O)

H(c)) = 1- (cos(2) - j.sin(2))+(cos(29) - j.sin(29))

Hem) 7-5.cos() +1 cos(29) + j(sin(n) - sin(2n))

Amplitude Response will be

|Hem) = 1 - cos(N) +cos(20) +; (sin(n) – įsin (29))|

|H(ej)= 1 (1 - E cos(n) + £cos(2n))* + (sin(n) – įsin(2n))"

(1 - cor(a) + cos(am) (1-5 cos(n)) +1(1-- cos(A)) cos(20) + 7.cos (20)

(1 - cos(n) + y cos(any) = 1 - 3 cosm) + 35 cos (M) + y cos(20) - cos(D) cos(22) + +.cos (20)

(sin(n) – asin(26)* = sin sin? (N) – 2 x= x=sin(m) sin(20) +2a sin’(21) 5 1 - 2x= x=sin(12) sin(21)+ 6 6 sin (21)

+ (1 - co(m) +acoul an ) + (-imen) -ásincan)" -1-cos() + 35 cos”(m) +şcos(21) - cos(n) cos(21) da.cos”(20) + 25 sin? (N) – 2 x= x=sin(m) sin(20) +sinº(21) L +

(1-cos(m) +.cos(20) + (6 sin(n) sin(20) + 5 25 cos(12) cos(21)+ sin(12) sin(21) L 18 +

(1-cos(m) +.cos(20) + (6 sin(n) sin(20) =1+3+ $cos(n) +şcos(21) (cos(2) cos(29) + sin(12) sin(20)]

(1 - cos(n) +.cos(20) + (7 sin(n) -+ sin(2n))* - cos(n) +şcos(21) (cos(1) (1 – 2 sinº(2)) + sin(12) 2 sin(1) cos(12)]

(1- cos(n) + cos2n)* + (sin(n) - h.sin(20) - 1 * cos(n) +.cos(21) (cos(1) – 2. sin’() cos(1) + 2 sin? (2) cos(2)]

(1- cos(n)+(cos(2n)* + (* sin(m) –sin(2n)) = cos(n) +şcos(21) - [cos()]

(1 - cos(m) + acos(2n) * + (6 sin(n) = sin(2n) 1 _2 cos(m) +_cos(20)

|H(ej)= 1 (1 - E cos(n) + £cos(2n))* + (sin(n) – įsin(2n))"

\H(239)|=- 3 cos(N) +{cos(21)

At

\H(e)")] = 7 21 hs cos(0) +şcos(0)

{ = = |(urə)

At

He") = 131 35 (I), I V

|H(en) = == 1.69688

At

2 = 27

|H(en) = - cos (9) + şcos (2x1)

|H(en) =

|H(en)= 31 ī=0.84853 18-3

At

|H(en) = 1 - cos () +cos (2x3)

(0) + (-) S - -= |(ure) |

|H(en) = 31 18 : = 0.56822 35 1872

At

\H(214)| = 7 33 cos(T) +} cos(2 x 76)

co_$+ sê " TE = |(412) S'O = I

Using these values we can plot the amplitude response as follows

Amplitude Response (ə) 0.5 1.5 2.5 2 (rad/sample)

Question (b)

f = 2 MHz= 2 X 106Hz

2πf Ω =

So

26

The frequency corresponding to rad/sample

s = f = 0 HZ

The frequency corresponding to rad/sample

х 2x 106 -= 250000 = 0.25 MHz 27

The frequency corresponding to rad/sample

2 X 106 -= 500000 = 0.5 MHz

The frequency corresponding to rad/sample

2. 3.1* 2 x 105 - = 750000 = 0.75 MHz 2006

The frequency corresponding to rad/sample

f = nf. пX 2 х 10° -= 1000000 = 1 MHz 2л 2п

So the plot will be

Amplitude Response 2.5 | yje) 0.5 1.5 0.5 MHz 0.75 MHz 2.5 3 1 MHz 3.5 2 (rad/sample) OH: 0.25 MHz

Question (d)

From the amplitude response, we can say that the given system is a low pass filter. This is because as frequency increases, the gain decreases.

The cut off frequency corresponds to the frequency when the gain is   times the maximum gain. Here the maximum gain is 3 at 0 rad/sample.

So the gain at cut off frequency will be

So

ဒီ ၌cos(0)+ဒ္ဒcos(20)

Squaring

11 - 1scos(0.) + y cos(20)

31 35 2-cos() + 3 cos(20) = 2

31 35 23 cos(1) + 3(2 cos(Q)-1) = 2

31 35 2-3 cos(1) + 6 cos(Q)-3= 2

6 cos(೧) - cos(೧) +3-5 =0

6 cos (9.) – 3 cos(_)+

Solving

cos(1) = 2.0721 cos(12)= 0.84455

First value cant be correct as cosine of a number can never be greater than 1

So the correct one is

cos(22) = 0.84455

So

1 = cos-(0.84455) = 0.56507 rad

So the cut off frequency in Hz will be

2.f. 27 0.56507 X 2 X 106 -= 179867.3674Hz = 0.1799 MHz 211