Olin College of Engineering
ENGR2410 – Signals and Systems
Lecture 2 Reference
Definitions
Linear system:
If
x 1 → L → y1
and
x 2 → L → y2
then
ax1 + bx2 → L → ay1 + by2 .
Time-invariant system:
If
x1 (t) → T I → y1 (t)
then
x1 (t − T ) → T I → y1 (t − T ).
Eigenfunctions
est → LT I →
H(s)
| {z }
est
|{z}
transfer function eigenfunction
Example:
If vin = est in the circuit above, when know that vout = H(s)est . Find H(s) by substituting
in the differential equation:
v¨o + 2αv˙o + ω02 vo = ω02 vi
st
st
st
st
H(s)s2
e
+ 2αH(s)s
e
+ ω02 H(s)
e
= ω02
e
H(s)(s2 + 2αs + ω02 ) = ω02
ω02
H(s) = 2
s + 2αs + ω02
Sinusoidal Steady State
A special (but important) case is when s = jω such that the input is ejωt , a complex
exponential. By Euler’s Equation,
ejωt = cos(ωt) + j sin(ωt).
We can “construct” the cosine function using a linear combination of two complex exponentials:
1
cos(ωt) = (ejωt + e−jωt ).
2
By linearity,
cos ωt → LT I → output
is equivalent to
1 jωt
1
e → LT I → H(jω)ejωt
2
2
+
+
1 −jωt
1
e
→ LT I → H(−jω)e−jωt
2
2
2
Therefore, the output is
1
1
H(jω)ejωt + H(−jω)e−jωt .
2
2
If H is holomorphic, then H(−jω) = H ∗ (jω), where the asterisk denotes the complex conjugate. Likely all the functions you will use will be holomorphic.
In polar form,
H(jω) = |H(jω)|ej∠H(jω)
H(−jω) = H ∗ (jω) = |H(jω)|e−j∠H(jω)
So the output of the system is
1
1
|H(jω)|ej∠H(jω) ejωt + |H(jω)|e−j∠H(jω) e−jωt
2
2
ej(ωt+∠H(jω) + e−j(ωt+∠H(jω)
=|H(jω)|
2
=|H(jω)| cos(ωt + ∠H(jω)).
The general solution for the sinusoidal steady state is
cos ωt → LT I → |H(jω)| cos(ωt + ∠H(jω))
Note:
(i) the output has the same frequency as the input
(ii) the output is scaled and shifted by H(jω)
For example, the transfer function of 2nd order system in the previous section is
H(jω) =
ω02
−ω 2 + j2αω + ω02
Its magnitude and phase are
(
ω02
|H(jω)| = √ 2
,
(ω0 − ω 2 )2 + (2αω)2
∠H(jω) = − tan
−1
2αω
2
ω0 − ω 2
Therefore, if
vi = V cos(ωt),
then
vo = V √
ω02
(ω02 − ω 2 )2 + (2αω)2
[
(
−1
cos ωt − tan
3
2αω
2
ω0 − ω 2
)]
.
)
.
Bode Plot
The plot of |H(jω)| and ∠H(jω) as a function of ω completely describes H(jω). A Bode
plot shows log |H(jω)| as a function of log ω for the magnitude and ∠H(jω) as a function of
log ω for the angle.
In order to sketch a Bode plot, we consider the asymptotes at both extremes. For example,
the transfer function of the 2nd order system in the previous section is
ω02
−ω 2 + j2αω + ω02
ω02
If ω → 0, then H(jω) ≈ 2 = 1
ω0
ω2
If ω → ∞, then H(jω) ≈ − 02
ω
H(jω) =
and
and
|H(jω)| = 1,
|H(jω)| =
ω02
,
ω2
∠H(jω) = 0
∠H(jω) = ±π
ω2
In the magnitude plot, the two lines intersect when 1 = ω02 , or ω = ω0 , the resonant frequency.
If we substitute the resonant frequency into the transfer function, we find
H(jω0 ) =
ω02
ω0
= −j
j2αω0
2α
and
|H(jω)| =
ω0
, Q,
2α
π
∠H(jω) = − .
2
The variable Q is called the quality factor and is widely used to characterize 2nd order
systems. The Bode plot is shown below and larger on the next page.
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