IName: I Exam 3 ECE301 Signals and Systems Thursday, April 18, 2019

Cover Sheet

Write your name on this page and every page to be safe. Test Duration: 80 minutes. Coverage: Comprehensive

Open Book but Closed Notes. Three two-sided handwritten sheets. Calculators NOT allowed.

This test contains two problems, each with multiple parts. You have to draw your own plots.

You must show all work for each problem to receive full credit.

·Problem i -Consider an analog sigrral- xa·(t) -below with- maximum frequency- wi.1 -= 4(.) ·­

rads/sec: the Fourier Transform, Xa(w), of Xa(t) is exactly zero for lwl > 40 rads/sec.

( ) { 7r sin ( 5 t) sin ( 15 t) } . . ( ) Xa t = - 2J sm 20t 5 Jrt Jrt

This is the same signal that is sampled for each and every part of entire Problem 1.

VIP: For EACH part, you are required to plot the DTFT of h(t) which is the lowpass filter impulse response that is used for interpolation and determine (and state) whether it is flat over the band -wM < w <WM and whether it filters out all the spectral replicas outside of the band -wM < w <WM, where WM= 40 rads/sec for all parts.

Problem 1 (a). The signal xa(t) is sampled at a rate Ws = 100 rads/sec., where Ws = 27r/Ts such the time between samples is Ts = {~ sec. This yields the discrete-time sequence

_ _ { 7r sin ( ~ n) sin ( ~~ n) } . . ( 27r ) x[n] - Xa(nTs) - - 2J sm -n 5 IrnTs IrnTs 5

27r where: T = -

s 100

A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal Xr(t). Show work.

n=-oo

where: T = 27r and h(t) =Ts!!_ sin(lOt) sin(50t) s 100 10 Jrt Jrt

00

Xr(t) = L x[n]h (t - nTs)

Problem 1 (b). Consider the SAME analog signal xa(t) with maximum frequency (band­width) WM = 40 rads/sec. This signal is sampled at the same rate Ws = 100 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr(t) = Xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = xa(t) or not.

00

where: T = 27r and h(t) = T ~ {sin(40t) + sin(60t)} s 100 8 2 Jrt Jrt Xr(t) = L x[n]h (t - nTs)

n=-oo

Problem 1 (c). Consider the SAME analog signal Xa(t) with maximum frequency (band­width) WM = 40 rads/sec. This signal is sampled at the same rate w8 = 100 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr ( t) = Xa ( t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = Xa(t) or not.

00

27r d () rl{sin(45t) sin(55t)} where: T = - an h t = - + ---s 100 s 2 Jrt Jrt Xr(t) = L x[n]h (t - nTs)

n=-oo

Show all your work for Prob. 1, parts (a)-(b)-(c) .

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Pmblem -1- tdj .- Consider -an analog signal Xa (-t )- wit-h- maximum -frequency (bandwidth] - -WM = 40 rads/sec. That is, the Fourier Transform of the analog signal xa(t) is exactly zero for fwf > 40 rads/sec. This signal is sampled at a rate Ws = 80 rads/sec., where Ws = 27r /Ts such the time between samples is Ts = ~~ sec. This yields the discrete-time sequence

{ . (7[ ) . (371" ) } 7r sm 8n sm 8 n 7r x[n] = Xa(nTs) = - T 2j sin (-n)

5 7rn s 7rnTs 2 where: T = 27r

s 80

A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal Xr(t). Show work.

00

Xr(t) = L x[n]h (t - nTs) 27r h(t) = Tssin(40t) where: Ts = 80 and 7rt n=-oo

Problem 1 ( e). Consider the SAME analog signal Xa ( t) with maximum frequency (band­width) WM = 40 rads/sec. This signal is sampled at the same rate Ws = 80 rads/sec., but reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr(t) = xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = Xa(t) or not.

00

where: T = 27r and h(t) = T { ~ sin(5t) sin( 40t)} s 80 s 5 7rt 7rt

Xr(t) = L x[n]h (t - nTs) n=-CXJ

Problem 1 (f). Consider the SAME analog signal xa(t) with maximum frequency (band­width) WM = 40 rads/sec. This signal is sampled at the same rate Ws = 80 rads/sec., but reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr(t) = xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = Xa(t) or not.

n=-oo

where: T = 27r and h(t) = T ~ {sin(40t) + sin(lOt)} s 80 s 2 7rt 7rt

00

Xr(t) = L x[n]h (t - nT3 )

Show all your work for Prob. 1, parts (d)-(e)-(f).

Show your work for Prob. 1, parts (d)-(e)-(f) .

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Problem 1 (g). Consider an analog signal Xa(t) with maximum frequency (bandwidth) WM= 40 rads/sec. This signal is sampled at a rate Ws =:= 5i) rads/_sec., where Ws_=_2_7r/Ts_

- sudi the tiine between sainpfos-is Ts-;;_ ;~ sec, yielding-the following discrete-time sequence:

{

• ( 7l" ) • ( 371" ) } 7r sm 5n sm 5 n 47r x[n] = Xa(nTs) = - 2j sin (-n)

5 7rnTs 7rnTs 5

27r where: T. = -

s 50

A reconstructed signal is formed from the samples above according to the formula below. Determine a closed-form expression for the reconstructed signal Xr(t). Show all work.

00

Xr(t) = L x[n]h (t - nTs) n=-oo

h T. 27r d h(t) __ T.ssin(25t) w ere: s = 50 an 7rt

Problem 1 (h). Consider the SAME analog signal Xa(t) with maximum frequency (band­width) WM = 40 rads/sec. This signal is sampled at the same rate Ws = 50 rads/sec., where Ws = 27r /Ts and the time between samples is Ts = ;~ sec, but at a different starting point. This yields the Discrete-Time x[n] signal below:

[ ] = ( T. 05T.)={~}{sin(%(n+0.5))sin(¥(n+0.5))} 2 ·. (47r( 05)) Xe n Xa n s + . s ( )T. ( ) J sm n + . 5 7r n + 0.5 s 7r n + 0.5 Ts 5

A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal xr(t). Hint: before you do a lot of work, look at the interpolating lowpass filter being used below.

n=-oo

27r sin(25t) where: Ts= - and h(t) =Ts---

50 7rt

00

Xr(t) = L Xe[n]h (t - (n + 0.5)Ts)

Show your work for Prob. 1, parts (g)-(h) .

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2

Problem 2. Consider the input signal x 0 (t) below.

Xo ( t) = e-j32t +e-12st +e-J24t +e-Jl6t +e-112t +e-jst +e-j4t + 1 +eJ4t +eJSt +ej12t +e116t +e124t +eJ:st +ej32t - - - -

- - - -

This signal is first input to an analog filter with impulse response

sin (5t) sin (5t) sin (5t) hLp(t) = 3 2j sin(5t) + 2 2j sin(15t) + 2j sin(25t)

Jrt Jrt Jrt

to form x(t) = x0 (t) * hLp(t), and then x(t) is sampled at a rate of W 8 = 64 to form x[n], so 2Jr

that the time between samples is Ts = 64 . The DT signal x[n] thus obtained is then input

to a DT LTI system with impulse response

(1)

Show all work. Write your expression for the output y[n] = x[n] * h[n] in the space below. Plot both the Fourier Transform of hLP(t) and the DTFT of h[n] to help solve the problem, but since the input is a sum of sinewaves, it is not necessary to plot the Fourier Transform of x0(t) or the DTFT of the sampled signal x[n].

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frequency (radians)

−80−70−60−50−40−30−20−10 0 10 20 30 40 50 60 70 80−4

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−2

−1

0

1

2

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4

frequency (radians)

Problem 2 y; . . ou can contmue your work for 2 h ere.

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