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EE102 Prof. S. Boyd
EE102 Homework 2, 3, and 4 Solutions
- Some convolution systems. Consider a convolution system,
y(t) =
∫ (^) +∞
−∞
u(t − τ )h(τ ) dτ,
where h is a function called the kernel or impulse response of the system.
(a) Suppose the input is a unit impulse function, i.e., u = δ. What is the output y? (This explains the terminology above.) (b) Suppose h = δ. What does the system do? (c) Suppose h is a unit step function. What does the system do? (d) Suppose h = δ′. What does the system do? (e) Suppose h(t) = δ(t − 1). What does the system do? (f) Suppose h is a rectangular pulse signal that is one between 0 and 1. What does the system do?
Solution: To solve these problems we just plug the given u or h into the convolution formula.
(a) If u = δ then we have
y(t) =
∫ (^) +∞
−∞
δ(t − τ )h(τ ) dτ = h(t).
In other words, the output y is the same as h. That’s why we call h the impulse response; it’s what comes out if an impulse goes in. (b) If we have h = δ, we find that y(t) = u(t). In words, this system does nothing: its output is exactly the same as its input. (c) If h is a unit step function, which we’ll denote h(t) = 1 (t) then we have
y(t) =
∫ (^) +∞
−∞
u(t − τ ) 1 (τ ) dτ =
∫ (^) ∞
0
u(t − τ ) dτ =
∫ (^) t
−∞
u(τ ) dτ.
(We changed variables in the last expression.) In words, the output is the integral of the input; this system is just an integrator. (d) If h = δ′, then we have
y(t) =
∫ (^) +∞
−∞
u(t − τ )δ′(τ ) dτ
Integrating by parts, we find
y(t) = u(t − τ )δ(τ )|+−∞∞ −
∫ (^) +∞
−∞
d dτ
(u(t − τ ))δ(τ ) dt =
∫ (^) +∞
−∞
u′(t − τ )δ(τ ) dt = u′(t)
Since y(t) = u′(t), this system is a differentiator.
(e) If h(t) = δ(t − 1) then the output is delayed, i.e., u(t − 1).
y(t) =
∫ u(t − τ )δ(τ − 1) dτ = u(t − 1).
This system is a 1-second delay. (f) If h is a rectangular pulse signal that is one between 0 and 1, then we have
y(t) =
∫ (^1)
0
u(t − τ ) dτ
Changing variables we can write this as
y(t) =
∫ (^) t
t− 1
u(τ ) dτ.
This shows that output is the integral (or in this case average) of the input over the last second.
- Describe the system shown below as an LCCODE. Hint: first label all signals, then write down how they are related.
PSfrag replacements ∫^ ∫
y
u
Solution: The signal going into the righthand integrator is y′, and similarly, the signal going into the lefthand integrator is y′′. The signal at the far left of the system, which is y′′, consists of the sum of three terms: αy′, βy, and u. We can write this out as an equation: y′′^ = αy′^ + βy + u. This can be expressed as the LCCODE
y′′^ − αy′^ − βy = u.
- Block diagram from equations. An interconnected set of systems is described by the following equations: v = A(u − v), w = B(v − z), z = C(w). Here u, v, w, z are signals and A, B, C are systems. You can consider u as the external input to the interconnected systems, and z as the external output of the interconnected system.
(a) Draw a pretty block diagram representing these equations. Hint: it usually takes two or three passes to get a pretty block diagram. (b) Now suppose that the systems A, B, C are scaling systems with gains a, b, c, respectively. Express z in terms of u. (In other words, eliminate the signals v and w from the equations.)
Solution. Starting from the output, we have:
y = cu + y − cu = cu + au + u + b(y − cu) = (c + a + 1 − bc)u + by
Solving the above equation for y, we obtain:
y =
c + a + 1 − bc 1 − b
u.
- Find the Laplace transform of the following functions.
(a) f (t) = (1 + t − t^2 )e−^3 t.
(b) f (t) =
0 0 ≤ t < 1 1 1 ≤ t < 2 − 1 2 ≤ t (c) f (t) = 1 − e−t/T^ where T > 0.
Solution: Finding F (s) is simply a matter of integration:
(a)
F (s) =
∫ (^) ∞
0
(1 + t − t^2 )e−^3 te−stdt
∫ (^) ∞
0
e−(s+3)t^ + te−(s+3)t^ − t^2 e−(s+3)tdt
s + 3
(s + 3)^2
(s + 3)^3
(b)
F (s) =
∫ (^2)
1
e−st^ +
∫ (^) ∞
2
−e−st
e−s s
2 e−^2 s s =
s
[ e−s^ − 2 e−^2 s
]
(c)
F (s) =
∫ (^) ∞
0
( 1 − e−t/T^
) e−stdt
∫ (^) ∞
0
( e−st^ − e−(s+1/T^ )tdt
) dt
s
T
sT + 1
These can also be solved using the Laplace Transform and its properties, and the same answers are obtained.
- Convolution and the Laplace transform.
(a) Evaluate h(t) = e−t^ ∗ e−^2 t^ using direct itegration. (These signals are not defined for t < 0.) (b) Find H, the Laplace transform of h, using the expression for h you found in part (a). (c) Verify that H is the product of the Laplace transforms of e−t^ and e−^2 t.
Solution:
(a) To find h(t) by using the definition of convolution:
e−t^ ∗ e−^2 t^ =
∫ (^) t
0
e−τ^ e−2(t−τ^ )dτ
= e−^2 t
∫ (^) t
0
eτ^ dτ
= e−^2 t(et^ − 1) = e−t^ − e−^2 t
(b) To find H, the Laplace Transform of h, use linearity and the Laplace Transform of the exponential: L(eat) =
s − a Hence, L(e−t^ − e−^2 t) =
s + 1
s + 2 (c) For this part, we are verifying that convolution in the time domain corresponds to multiplication in the s-domain. In other words, in part (a) and (b) we convolved two signals, and then took the Laplace transform of the resulting signal. We want to show this is the same thing as taking the Laplace Transform of each signal, and then multiplying them. (L(e−t))(L(e−^2 t)) = (
s + 1
s + 2
Which is in fact equal to the answer in part (b). These signals are sketched below.
above is s^3 Y (s) − 2 sY (s) = U (s). Solve for Y (s) to get
Y (s) =
U (s) s^3 − 2 s
- Solve the following LCCODEs using Laplace transforms. Verify that the solution you find satisfies the initial conditions and the differential equation.
(a) dv/dt = − 2 v + 3, v(0) = −1. (b) d^2 i/dt^2 + 9i = 0, i(0) = 1, di/dt(0) = 0.
Solution: We will use the differentiation theorem to take the Laplace transforms. Note that this step reduces differential equations in time to algebraic equations in the s-domain:
(a)
dv/dt = − 2 v + 3
sV (s) − v(0) = − 2 V (s) +
s V (s) =
3 − s (s)(s + 2)
=
s
s + 2
thus v(t) = 32 − 52 e−^2 t. (b)
d^2 i/dt^2 + 9i = 0
s^2 I(s) − si(0) −
di dt
(0) + 9I(s) = 0
I(s) =
s s^2 + 3^2
thus i(t) = cos 3t.
- Four signals a, b, c, and d are related by the differential equations
a′^ + a = b, b′^ + b = c, c′^ + c = d,
where a(0) = b(0) = c(0) = 0. Express A(s), the Laplace transform of a, in terms of D(s), the Laplace transform of d. Solution. There are several ways to solve this. First Method Starting with the equation for d, we have
d = c′^ + c = b′′^ + b′^ + b′^ + b = a′′′^ + a′′^ + 2a′′^ + 2a′^ + a′^ + a = a′′′^ + 3a′′^ + 3a′^ + a (1)
Now, b′(0) = c(0)−b(0) = 0, a′(0) = b(0)−a(0) = 0, and, consequently, a′′(0) = b′(0)−a′(0) =
- So, all of the initial conditions that we need to worry about (i.e., a′′(0), a′(0), a(0)) are zero. Then, taking the Laplace transform of both sides of (1), we obtain
D(s) = s^3 A(s) + 3s^2 A(s) + 3sA(s) + A(s)
Solving for A(s) produces
A(s) =
D(s) s^3 + 3s^2 + 3s + 1 Second Method Since a(0) = b(0) = c(0) = 0, taking the Laplace transform of the three given equations produces
sA(s) + A(s) = B(s), sB(s) + B(s) = C(s), sC(s) + C(s) = D(s) (2)
Then, starting with the Laplace transform for D(s), we obtain
D(s) = sC(s) + C(s) = s^2 B(s) + sB(s) + sB(s) + B(s) = s^3 A(s) + s^2 A(s) + 2s^2 A(s) + 2sA(s) + sA(s) + A(s) = s^3 A(s) + 3s^2 A(s) + 3sA(s) + A(s)
So, as before, A(s) =
D(s) s^3 + 3s^2 + 3s + 1 Yet Another Method From the equations (2), we first write them as
(s + 1)A(s) = B(s), (s + 1)B(s) = C(s), (s + 1)C(s) = D(s),
and now we express this as
A(s) =
B(s) s + 1
, B(s) =
C(s) s + 1
, C(s) =
D(s) s + 1
Now, combining these, we get
A(s) =
s + 1
B(s) =
(s + 1)^2
C(s) =
(s + 1)^3
D(s),
which is the same answer.
- In the system shown below, k is a gain.
PSfrag replacements
∫ ∫ k
y
(a) Consider a metal wire, for which c > 0. If the current i is smaller than a critical value icrit, the temperature T converges to a steady-state value as t → ∞. If the current i is larger than this critical value of current, then the temperature T converges to ∞ as t → ∞. (In practice, the temperature increases until the conductor is destroyed, e.g., melted.) This phenomenon is called thermal runaway. Find the critical value icrit, above which thermal runaway occurs. Express the answer in terms of the other constants in the problem (a, b, R 0 , c). (b) Suppose the wire is initially at ambient temperature, i.e., T (0) = 0, and the constants have the values
a = 1J/◦C, b = 0.5W/◦C, i = 10A, R 0 = 1Ω, c = 0. 01 /◦C
Find T (t) for t ≥ 0
Solution:
(a) We just combine the two given equations to get
aT ′^ = −bT + i^2 R 0 (1 + cT ),
which we can write as T ′^ =
−b + i^2 R 0 c a
T +
i^2 R 0 a
This simple first order equation is stable only if the coefficient (−b+i^2 R 0 c)/a is negative, i.e., only if |i| ≤
√ b/(R 0 c) = icrit.
You can solve the ODE to verify that if |i| ≥ icrit, the temperature goes to ∞; for |i| < icrit, the temperature converges to some steady-state value. (b) Plugging in the values given, our differential equation becomes T ′^ = 0. 5 T + 100. We can solve this via Laplace transforms. Let Y denote the Laplace transform of T (which, unfortunately, is already capitalized!):
sY = 0. 5 Y + 100/s,
so Y =
s(s − 0 .5)
s
s − 0. 5
Therefore we have y(t) = 200
( e^0.^5 t^ − 1
) .
Note that in this case i is above the critical current; we have thermal runaway here.
- A voltage v(t) is applied to a DC motor. A simple electrical model of the motor is an inductance L in series with a resistance R, so the motor current i(t) satisfies
Ldi/dt + Ri = v.
The motor shaft angle is denoted θ(t), and the shaft angular velocity ω(t) (so we have ω = dθ/dt). The motor current puts a torque on the shaft equal to ki(t), where k is the motor
constant. The shaft rotational inertia is J and the damping coefficient is b. Newton’s equation is then: Jdω/dt = ki − bω.
Assuming that i(0) = 0, θ(0) = 0, and ω(0) = 0, express Θ (the Laplace transform of θ) in terms of V (the Laplace transform of v). The numbers L, R, k, J, b are all positive constants. Solution: Everything you need to know is in the three equations
Ldi/dt + Ri = v, ω = dθ/dt, Jdω/dt = ki − bω.
Let’s take the Laplace transform of these equations to get the algebraic equations
L(sI − i(0)) + RI = V, Ω = sΘ − θ(0), J(sΩ − ω(0)) = kI − bΩ.
From the assumption that i(0) = 0, θ(0) = 0, and ω(0) = 0, these equations simplify to
LsI + RI = V, Ω = sΘ, JsΩ = kI − bΩ.
Now we solve these equations for Θ in terms of V (and the constants). From the first equation we get I =
V
Ls + R
From the third equation we get Ω =
kI Js + b
Combining these we get Ω =
kV (Js + b)(Ls + R)
Combining this with the middle equation above yields
kV s(Js + b)(Ls + R)
- The waveform shown below is the current in a series RLC circuit. The value of the resistor is 100Ω.
(c)
(s − 2)(s − 3)(s − 4) s^4 − 1
Solution:
(a) This one is already in partial fraction form. We can see immediately the poles are −1, −2, and −3. From the partial fraction expansion we can just read off the inverse Laplace transform: e−t^ + e−^2 t^ + e−^3 t. To find the zeros and real factored form we’ll put everything on a common denominator: 1 s + 1
s + 2
s + 3
(s + 2)(s + 3) + (s + 1)(s + 3) + (s + 1)(s + 2) (s + 1)(s + 2)(s + 3)
=
3 s^2 + 12s + 11 (s + 1)(s + 2)(s + 3) To find the zeros we apply the quadratic formula to the numerator. This yields zeros at s = − 2 ±
√ 3
- These zeros are complex, so the last expression above is the real factored form. (b) We cannot factor the numerator over the reals (i.e., it has complex roots), and the denominator can be written as s(s + 1)(s − 1). This gives us the real factored form: s^2 + 1 s(s + 1)(s − 1)
The zeros are ±j, and the poles are 0, ±1: s^2 + 1 s^3 − s
(s + j)(s − j) s(s + 1)(s − 1) Performing the partial fraction expansion, we find s^2 + 1 s^3 − s
s
s − 1
s + 1 Thus the inverse Laplace transform is −1 + et^ + e−t. (c) The numerator is already in real factored form; the zeros are 2, 3, and 4. Now let’s factor the denominator. In fact there is an explicit but very ugly formula that gives all four roots of a general quartic (just like the quadratic formula), but we won’t need it here. (And by the way, there is no such formula for the roots of polynomials of degree five and higher!) To factor it we just notice that 1 and −1 are roots, or notice that we can write s^4 − 1 as (s^2 − 1)(s^2 + 1), and we can factor both terms. Either way we end up with s^4 − 1 = (s + 1)(s − 1)(s^2 + 1). This is as far as we can go over the reals. So the real factored form is (s − 2)(s − 3)(s − 4) (s + 1)(s − 1)(s^2 + 1)
The denominator factors further to (s + 1)(s − 1)(s + j)(s − j). Thus the poles are ± 1 and ±j. The partial fraction expansion is 15 s + 1
s − 1
− 6 .25 + 3. 75 j s + j
− 6. 25 − 3. 75 j s − j hence the inverse Laplace transform is 15 e−t^ − 1. 5 et^ − 12 .5 cos t + 7.5 sin t.
