Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Answers to Pre Lab Exercises - Laboratory | ECE 1563, Lab Reports of Electrical and Electronics Engineering

Material Type: Lab; Class: LABORATORY; Subject: Electrical and Computer Engineeri; University: University of Pittsburgh; Term: Spring 2009;

Typology: Lab Reports

Pre 2010

Uploaded on 09/02/2009

koofers-user-l4q
koofers-user-l4q 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ECE 1563 Spring, 2009
Answers to Pre-lab Exercises
Lab 1
1. Derive the impulse response, step response and transfer function for an RC (i.e., first-order) low-pass
filter described by the equation
)t(x
RC
1
)t(y
RC
1
dt
dy =+
Show a circuit diagram.
See slides 12-14 of the Introductory Lecture. Step response is the integral of the impulse
response: )t(u)e1(
RC
1
)t(h RC/t
u
=
2. Sketch the step response for filters with RC = 10 millisecond (ms) and RC = 100 ms. Does
increasing R make the system response more or less sluggish? Increasing R makes the response
more sluggish.
3. Find and sketch the frequency response of the filter for RC = 10 ms and for RC = 50 ms. A Bode
plot is adequate. Plot in slide is for RC = 333 ms. Since these are bode plots, the break point, at
1/RC, just moves to the right from that shown (from 3 to 20 for 50 ms and to 100 for 10 ms).
4. Find the convolution (analytically) of a pulse p(t) = u(t) – u(t-0.01), t in seconds, a) with an
impulse function and b) with itself.
Convolution of any function with an impulse function is just the function itself. Convolution of
a pulse with itself is a triangular function of twice the width of the pulse.
5. Prove the relation:
1for
1
1N
1N
0
nα
α
α
=α
directly. follows formula above The .α1α)αα(α as written becan which
αααα αααα)α1(Consider
NN
1N
1
nn0
1N
0
1N
0
1N
0
1N
0
1N
0
N
1
nn1nnnn
1N
0
n
=+
===
∑∑
−−
+
What is
N.
=αα
1N
0
n1? if
Lab 2
1. Find the exponential Fourier series coefficients of the following continuous-time, periodic signals:
sin 2π1000t t1000π2j
1
t1000π2j
1
t1000π2jt1000π2j eaeae
2j
1
e
2j
1
+==
sin 2π500t + sin 2π1000t
t1000π2j
2
t1000π2j
2
t500π2j
1
t500π2j
1
t1000π2jt1000π2jt500π2jt500π2j eaeaeaeae
2j
1
e
2j
1
e
2j
1
e
2j
1
+++=+=
(sin 2π500t)(sin 2π1000t) = ½ cos 2π500t - ½ cos 2π1500t
t1500π2j
3
t1500π2j
3
t500π2j
1
t500π2j
1
t1500π2jt1500π2jt500π2jt500π2j eaeaeaeae
4
1
e
4
1
e
4
1
e
4
1
+++=+=
pf3

Partial preview of the text

Download Answers to Pre Lab Exercises - Laboratory | ECE 1563 and more Lab Reports Electrical and Electronics Engineering in PDF only on Docsity!

ECE 1563 Spring, 2009

Answers to Pre-lab Exercises

Lab 1

1. Derive the impulse response, step response and transfer function for an RC (i.e., first-order) low-pass

filter described by the equation

x(t )

RC

y(t)

RC

dt

dy

Show a circuit diagram.

See slides 12-14 of the Introductory Lecture. Step response is the integral of the impulse

response: ( 1 e )u(t)

RC

h (t) t/RC u

− = −

2. Sketch the step response for filters with RC = 10 millisecond (ms) and RC = 100 ms. Does

increasing R make the system response more or less sluggish? Increasing R makes the response

more sluggish.

3. Find and sketch the frequency response of the filter for RC = 10 ms and for RC = 50 ms. A Bode

plot is adequate. Plot in slide is for RC = 333 ms. Since these are bode plots, the break point, at

1/RC, just moves to the right from that shown (from 3 to 20 for 50 ms and to 100 for 10 ms).

4. Find the convolution (analytically) of a pulse p(t) = u(t) – u(t-0.01), t in seconds, a) with an

impulse function and b) with itself.

Convolution of any function with an impulse function is just the function itself. Convolution of

a pulse with itself is a triangular function of twice the width of the pulse.

5. Prove the relation:

for 1 1

N 1 N

0

n α≠ −α

−α ∑ α =

whichcanbewrittenas α (α α ) α 1 α .Theaboveformulafollows directly.

Consider ( 1 α) α α α α α α α α

N N

N 1

1

0 n n

N 1

0

N 1

0

N 1

0

N 1

0

N 1

0

N

1

n n n n 1 n n

N 1

0

n

− − − − −

What is ∑ α α= N.

N− 1

0

n if 1?

Lab 2

1. Find the exponential Fourier series coefficients of the following continuous-time, periodic signals:

sin 2π1000t

j 2 π 1000 t 1

j 2 π 1000 t 1

j 2 π 1000 t j 2 π 1000 t e ae a e j 2

e j 2

− = − = +

sin 2π500t + sin 2π1000t

j 2 π 1000 t 2

j 2 π 1000 t 2

j 2 π 500 t 1

j 2 π 500 t 1

j 2 π 500 t j 2 π 500 t j 2 π 1000 t j 2 π 1000 t e ae a e a e a e j 2

e j 2

e j 2

e j 2

− −

− − = − + − = + + +

(sin 2π500t)(sin 2π1000t) = ½ cos 2π500t - ½ cos 2π1500t

j 2 π 1500 t 3

j 2 π 1500 t 3

j 2 π 500 t 1

j 2 π 500 t 1

j 2 π 500 t j 2 π 500 t j 2 π 1500 t j 2 π 1500 t e ae a e a e a e 4

e 4

e 4

e 4

− −

− − = + − − = + + +

EE 1563 Answers to pre-lab exercises Spring, 2009 page 2

2. Find the continuous Fourier series coefficients of symmetric and asymmetric continuous-time,

periodic square waves with period of two milliseconds and 50% duty cycle:

Wave 1 (symmetric):

x 1 (t) = 1 for -0.5 < t < 0.5, x 1 (t) = 0 for 0.5 < t < 1.

Wave 2 (asymmetric):

x 2 (t) = 1 for 0 < t < 1, x 2 (t) = 0 for 1 < t < 2

where times are in milliseconds.

jkπ/ 2 2 k

1 k

e πk

sinπk/ 2 Forx :a

πk

sinπk/ 2 Forx :a

3. Find analytically the discrete time Fourier transform (DTFT) of the sequence

x[n] = δ[n] + δ[n-1] + δ[n-2] + δ[n-3] + δ[n-4].

X

e ( 1 2 cosω 2 cos 2 ω )

(e ) x(t)e 1 e e e e e (e e 1 e e )

j 2 ω

jω j 2 ω j 3 ω j 4 ω j 2 ω j 2 ω jω jω j 2 ω

4

n 0

jω jωn

− − − − − − −

=

Normalize the sequence to make the maximum magnitude of the DTFT equal to 1.

Maximum magnitude is 5 at ω = 0, so the sequence should be divided by 5 to make the

maximum magnitude = 1.

4. Find an analytical expression for the frequency response of the discrete approximation to the low-

pass RC filter that you derived for Experiment 1 (Section 1) and sketch its magnitude and phase vs.

frequency.

The forward difference equation is y[n] = (1-aT) y[n-1] + aT x[n-1], where a = 1/RC.

e ( 1 T )

aT | 1 ( 1 aT)z

aTz H( e ) H(z)| 1 ze jω

1

ze

j ω jω jω − −

− =

= a

Lab 3

1. If a continuous-time signal x(t) has a Nyquist rate (the minimum sampling frequency required by the

sampling theorem to capture the signal) of 500 samples per second, what sampling frequency (in

samples per second) would be needed to capture the signal x(t) cos(1000πt)?

Maximum frequency in x(t) must be 250 Hz. The product x(t) cos(1000πt) will have sum and

difference frequencies (750 and 1250 Hz), so the sampling frequency must be 2x1250 or 2500

Hz.

2. You want to sample a 550 Hz symmetric square wave (zero-mean) in order to accurately capture at

least 95% of the total power. What sampling rate should you use?

Total power in the square wave is A

2

/4, where A/2 is the zero-to-peak amplitude. Power in the

harmonics is given by Parseval’s relation:

k 1

k k

k forkodd.

k π

A

2 |a | wherea 0 forkevenanda j

Note that the dc term is zero (zero mean).

The ratio of power in harmonics 1 through K to the total power is.

k

π

k π

A

A / 4

1 K

k 1

K

k 1

2 2

2 2

= =