# Solutions - Olin College

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```Franklin W. Olin College of Engineering
Needham, Massachusetts
Electrical and Computer Engineering
ENGR2410 – Signals and Systems
Assignment 2
Problem 1: Consider the RC circuit shown below.
A. Find vout (t) when vin = V sin ωt. Assume the system is in sinusoidal steady state (i.e.,
all transients have disappeared).
Solution:
Step 1: Find the ODE
vin = vR + vout
vin = RC vout
˙ + vout
1
1
vin
vout
˙ + RC vout = RC
Let RC = τ so that
vout
˙ + τ1 vout = τ1 vin
Step 2: Find the transfer function
Assume vin = ejωt find H(jω) such that vout = H(jω)ejωt
jωH(jω)ejωt + τ1 H(jω)ejωt = τ1 ejωt
1/τ
1/τ
H(jω) =
so |H(jω)| = √
and ∠H(jω) = − tan−1 (ωτ )
2
jω + 1/τ
ω + 1/τ 2
Step 3: Write solution
vout = |H(jω)|V sin(ωt + ∠H(jω))
or
vout = V √
1/τ
ω2
+ 1/τ 2
sin[ωt − tan−1 (ωτ )]
B. Assume an input vin = V sin(ωt)u(t) so that the circuit is initially at rest. Find an
expression for vout (t) when t &gt; 0.
Solution:
Step 1: Find particular solution
The sinusoidal steady state from the previous part is the particular solution for t &gt; 0.
Step 2: Find homogeneous solution
Solve vout
˙ + τ1 vout = 0
t
Integrate to obtain vout,homogeneous = Ae− τ
Step 3: Add the particular solution to the homogeneous solution and use the initial
condition to determine the constant of integration A
1/τ
t
vout = V √
sin[ωt − tan−1 (ωτ )] + Ae− τ , t &gt; 0
2
2
ω + 1/τ
vout (0) = 0
1/τ
sin[tan−1 (ωτ )] + A
0 = −V √
2
ω + 1/τ 2
1/τ
ω
√
A=V√
2
2
2
ω + 1/τ
ω + 1/τ 2
)
(
1/τ
ω
t
vout = V √
sin[ωt − tan−1 (ωτ )] + √
e− τ , t &gt; 0
2
2
2
2
ω + 1/τ
ω + 1/τ
2
C. Plot the solution assuming V = 1, ω = 1, and RC = 1 as well as V = 1, ω = 10, and
RC = 1.
Solution:
vout (ω=1)
1
0.5
0
−0.5
−1
0
1
2
3
0
1
2
3
4
5
6
7
4
5
6
7
vout (ω=10)
0.2
0.1
0
−0.1
−0.2
time (s)
3
Problem 2: Consider the RLC circuit shown below.
A. Find a differential equation that relates vin and vout .
Solution:
∫
vin − vout
vout
= C v̇out +
dt
R
L
v̇in − v̇out
vout
= v̈out +
RC
LC
v̇out vout
v̇in
v̈out +
+
=
RC
LC
RC
2
v̈out + 2αv̇out + ω0 vout = 2αv̇in
α=
1
2RC
1
w0 = √
LC
B. Derive an expression for the transfer function from vin to vout .
Solution:
Assume vin = ejωt find H(jω) such that vout = H(jω)ejωt
v̈out + 2αv̇out + ω02 vout = 2αv̇in
1
1
w0 = √
α=
2RC
LC
2
jωt
− ω H(jω)e + 2αjωH(jω)ejωt + ω02 H(jω)ejωt = 2αjωejωt
H(jω) =
−ω 2
2αjω
+ 2αjω + ω02
4
C. Sketch the Bode plot for the transfer function from vin to vout using asymptotic approximations.
Solution:
2αjω
+ 2αjω + ω02
2αjω
If ω → 0, then H(jω) ≈
ω02
2αω
|H(jω)| = 2 , ∠H(jω) = π/2
ω0
2αjω
2αj
If ω → ∞, then H(jω) ≈ − 2 = −
ω
ω
2α
|H(jω)| =
, ∠H(jω) = −π/2
ω
2αω
2α
Intersection
=
⇒ ω = ω0
2
ω0
ω
2α
1
Asymptote value at intersection
=
ω0
Q
but the actual value is
H(jω0 ) = 1
|H(jω0 )| = 1, ∠H(jω0 ) = 0
H(jω) =
−ω 2
The magnitude of 1/Q determines whether the system resonates or not. If Q ≪ 1,
then α ≫ ω0 . This system is overdamped and does not oscillates. It behaves as two
first order systems together.
0
|H(jω)|
10
−2
10
−4
10
−3
10
−2
10
−1
0
10
10
1
10
2
10
3
10
∠ H(jω)/π
0.5
0
−0.5
−3
10
−2
10
−1
10
0
10
frequency (ω)
5
1
10
2
10
3
10
If Q ≫ 1, then α ≪ ω0 . This system is underdamped and resonates.
0
|H(jω)|
10
−2
10
−4
10
−3
10
−2
10
−1
0
10
10
1
10
2
10
3
10
∠ H(jω)/π
0.5
0
−0.5
−3
10
−2
10
−1
10
0
10
frequency (ω)
6
1
10
2
10
3
10
```