Signals and Systems Assignment Help

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1. Second-order systems
The impulse response of a second-order CT system has the form
where the parameters σ, ωd, and φ are related to the parameters of the characteristic
polynomial for the system: s2 + Bs + C.
a. Determine expressions for σ and ωd (not φ) in terms of B and C.
Express the impulse response in terms of complex exponentials:
The impulse response is a weighted sum of modes of the form
where s0 and The characteristic polynomial has the form
Thus B = 2σ and C = σ2 + Solving, we find that
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b. Determine
– the time required for the envelope of h(t) to diminish by a factor of e,
– the period of the oscillations in h(t), and
– the number of periods of oscillation before h(t) diminishes by a factor of e.
Express your results as functions of B and C only.
The time to decay by a factor of e is
The period is
The number of periods before diminishing a factor of e is
Notice that this last answer is equivalent to Q/π where Q =

c. Estimate the parameters in part b for a CT system with the following poles:
From the plot σ = 10 and ωd = 100.
The time to decay by a factor of e is 0.1
The period is
The number of cycles before decaying by e is
The unit-sample response of a second-order DT system has the form

where the parameters r0, Ω0, and Φ are related to the parameters of the characteristic
polynomial for the system:
d. Determine expressions for r0 and Ω0 (not Φ) in terms of D and E.
Express the unit-sample response in terms of complex exponentials:
The poles have the form z = r0e jΩ0 and z = r0e−jΩ0 . The characteristic
equation is z2 + Dz +E = (z −r0e jΩ0 )(z −r0e−jΩ0 ) = z2 −2r0 cos Ω0 +r2 0.
Thus D = −2r0 cos Ω0 and E = r2 0. Solving, we find that
e. Determine
- the length of time required for the envelope to diminish by a factor of e.
– the period of the oscillations in h[n], and

- the number of periods of oscillation in h[n] before it diminishes by a factor of e.
Express your results as functions of D and E only.
The time to diminish by a factor of e is Taking the log of both sides yields
n ln r0 = −1 so that the time is
The period is 2π/Ω0 which is
The number of periods before the response diminishes by e is
f. Estimate the parameters in part e for a DT system with the following poles:

From the plot Ω0 = tan−1
The time to decay by a factor of e is
The number of cycles before decaying by e is
2. Matches
The following plots show pole-zero diagrams, impulse responses, Bode magnitude
plots, and Bode angle plots for six causal CT LTI systems. Determine which
corresponds to which and fill in the following table.

Pole-zero diagram 1 has a single pole at zero. The impulse response of a system with
a single pole at zero is a unit step function (3). We evaluate the frequency response by
considering frequencies along the jω axis. As we move away from the pole at the
origin the log-magnitude decays linearly (5). The phase is constant since the angle
between the pole and any point along positive side of the jω axis remains constant at
π/2. The angle of the frequency response is therefore −π/2 (4).
Pole-zero diagram 4 has a single pole at at s = −1. The impulse response has the form
estu(t) = e−t u(t) (2). As we move along the jω axis, we move away from the pole at
the origin, and the log-magnitude will eventually decay linearly. Because the pole is
not exactly at the origin, this decay is not significant until ω = 1 (6). The phase starts at
0, and eventually moves to −π/2. Note that as we move farther up the jω axis, this
system behaves like the system of diagram 1 (2)
Pole-zero diagram 3 adds a zero at the origin. A zero at the origin corresponds to
taking the derivative, so we take the impulse response of pole-zero diagram 4 (2) and
take its derivative (4). When ω is small, the zero is dominant. As we move away from
ω = 0, the effect of the zero diminishes and the log-magnitude increases linearly. For
sufficiently large ω we are far enough that the zero and pole appear to cancel each
other, and the magnitude becomes a constant (3). A zero at the origin means that we
take the phase response of pole-zero diagram 4 (2) and add π/2 to it (6).

Pole-zero diagram 2 contains complex conjugate poles
The impulse response has the form
which is response (1). The magnitude response will eventually decay twice as
fast as that of pole-zero diagram 4 (6). Since there are two poles, there will be a
bump at around ω = 1 (2). At the origin, the angular contributions of the two
poles cancel each other out, hence the angle is zero. As we move up the jω axis,
the angles add up to −π, with each pole contributing −π/2 (3).
Pole-zero diagram 6 adds a zero at the origin, meaning that we take the
derivative of the impulse response of pole-zero diagram 2 (1). The derivative
ends up being the combination of a decaying cos(t) term minus a decaying sin(t)
term (5). The zero at the origin adds a linearly increasing component to the
magnitude function (4). It also adds π/2 to the phase response everywhere (5).
Pole-zero diagram 5 has complex conjugate poles and zeros at the same
frequency ω. The system function has the form

This denominator has the same form as pole-zero diagrams 2 and 6, but has an additional
power of s (corresponding to differentiation) in the numerator. This leads to a response of
the form in (6). The symmetry of the poles and zeros means they cancel each other’s effect
on magnitude (1). The phase response at ω = 0 is zero, as the contributions cancel each
other out. As we move past ω = 1 where the conjugates are located, the phase moves in the
negative direction faster, but eventually settles back at 0 as we move farther and the
contributions again cancel each other out (1)

Engineering Design Problems
3. Desired oscillations
The following feedback circuit was the basis of Hewlett and Packard’s founding
patent.
a. With R = 1 kΩ and C = 1µF, sketch the pole locations as the gain K varies from
0 to ∞, showing the scale for the real and imaginary axes. Find the K for which the
system is barely stable and label your sketch with that information. What is the
system’s oscillation period for this K?
The closed-loop gain is
The denominator is zero if

The point of marginal stability is where the root locus crosses the jω axis. This
occurs when the real part of −1 +

so that K = 8. The frequency of oscillation is ω = so the period of oscillation is
For RC = 1 ms (as given), the period T = 3.63 ms.
b. How do your results change if R is increased to 10 kΩ?
Increasing R by a factor of 10 increases the period T by a factor of 10, to T = 36.3
ms. It has no effect of the critcal value of K = 8.
4. Robotic steering
Design a steering controller for a car that is moving forward with constant velocity V.

You can control the steering-wheel angle w(t), which causes the angle θ(t) of the car
to change according to
where d is a constant with dimensions of length. As the car moves, the transverse
position p(t) of the car changes according to
Consider three control schemes:
a. w(t) = Ke(t)
b. w(t) = Kve˙(t)
c. w(t) = Ke(t) + Kve˙(t)
where e(t) represents the difference between the desired transverse position x(t) =
0 and the current transverse position p(t). Describe the behaviors that result for
each control scheme when the car starts with a non-zero angle (θ(0) = θ0 and
p(0) = 0). Determine the most acceptable value(s) of K and/or Kv for each
control scheme or explain why none are acceptable.

Part a. This system can be represented by the following block diagram:
We are given a set of initial conditions — p(0) = 0 and θ(0) = θ0 — and we are asked
to characterize the response p(t). Initial conditions are easy to take into account when
a system is described by differential equations. However, feedback is easiest to
analyze for systems expressed as operators or (equivalently) Laplace transforms.
Therefore we first calculate the closed-loop system function,
which has two poles: ±jω0 where ω0 = V We can convert the system function
to a differential equation:
and then find the solution when x(t) = 0,

so that p(t) = C sin ω0t since p(0) = 0.
From p(t) we can calculate θ(t) = ˙p(t)/V = C/V ω0 cos ω0t. From the initial condition
θ(0) = θ0, it follows that C = V θ0/ω0 and
for t > 0.
If K is small, then the oscillations are slow, but they have a large amplitude. If K is
large, then the oscillations are fast (and therefore uncomfortable for passengers), but the
amplitude is small. While none of these behaviors are desireable, it would probably be
best to increase K so that the amplitude of the oscillation is small enough so that the car
stays in its lane.
Part b. The system can be represented by the following block diagram:
The closed-loop system function is

The closed-loop poles are at s = 0 and s =
Since p(0) = 0, the form of p(t) is given by
for t > 0. We can find C by relating C to the initial value of θ(t) = ˙p(t)/V . Since θ(0) =
θ0, p˙(0) = V θ0. Therefore C =
for t > 0 as shown below.

We would like to make Kv large because large Kv leads to fast convergence. Large values
of Kv also lead to smaller steady-state errors in p(t).
There are no oscillations in p(t) with the velocity sensor, which is an advantage over
results with the position sensor in part a. However, there is now a steady-state error in
p(t), which is worse. Fortunately the steady-state error can be made small with large Kv.
Part c. The system can be represented by the following block diagram:
The closed-loop system function is
This second-order system has a resonant frequency ω0 = and a quality factor

There is an enormous variety of acceptable solutions to this problem, since there are
many values of K and Kv that can work. Here, we focus on one line of reasoning
based on our normalization of second-order system in terms of Q and ω0.
To avoid excessive oscillations, we would like Q to be small. Try Q = 1. Then
Then p(t) has the form
As before, we can use the intial condition of θ(0) = θ0 to determine C. In general,
θ(t) = ˙p(t)/V so p˙(0) = θ0V = C √ 3ω0/2. Therefore

Increasing Q would reduce the overshoot but slow the response. We could compensate
for the slowing of the response by increasing ω0.
Performance can be adjusted to be better than either part a or part b. By adjusting Q and
ω0 we can get convergence of p(t) to zero with minimum oscillation.
Although the steady-state value of the error is zero and the oscillation is minimized,
there is still a transient behavior, which could momentarily move the car into the other
lane

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