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
n≠α
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−
−
−
−
−− +++=−−+=
