For the three circuits shown:

Circuit i | |

Circuit ii | |

Circuit iii |

a) Write the three differential equations (input=f(t)=ei(t), output=y(t)=eo(t)).
in standard form:
where a, b, and d are constants and the right hand side may have derivatives of the input
(f(t))as well.

Hint: First write a loop equation for in terms of i(t)
(current), and then solve.

For the next parts assume the initial conditions at t=0^{-} are all 0 (i.e., a zero-state
problem), and the input is a unit step (f(t)=γ(t)).
Also assume that the system is overdamped so there are no oscillations in
the homogeneous response.

b) Find the output of each system as t→∞. This is
easily done if you assume the particular solution is a constant (y_{p}(t)=K).

c) Find initial
conditions on y(t) and dy(t)/dt at t=0^{+}. Again, start with
i(0^{+}) and its derivative. One of them is a bit tricky.

d) Very crudely sketch the step response for an overdamped system. Pay particular attention to the response for t near 0 and ∞. Note there are no actual numbers so you can't sketch with any accuracy.

e) Answer the following questions. Based on the order of the derivative of the input...

- ...what can you say about the steady state value of the output as t→∞?
- ...what can you say about the nature of any
discontinuities of the system between t=0
^{-}and t=0^{+}.

a)

ai)

aii)

aiii)

b) As t→∞ the output goes to zero if the input is differentiated (as t→∞ the right side is zero if there is a derivative, and this makes the particular response zero).

c) At t=0+ there is no current, so i(0+) is 0.

e_{i}=e_{R}+e_{C}+e_{L}. At t=0^{+}
the e_{i}(0^{+})=1, e_{R}(0^{+})=0 (i=0), e_{C}(0^{+})=0
(voltage across capacitor can't change instantaneously), so e_{L}(0^{+})=1
(di/dt = e_{L}/L =1/L at t=0^{+})

ci)

cii)

ciii)

d) Note each output looks like the derivative of the one that precedes it.

e) If there is any derivative of the input, the output goes to zero as
t→∞. If the order of the derivative on the right side is
the same as the order of the highest derivative on the left side, then there
is a discontinuity of the output between t=0^{-} and t=0^{+}.
(Note: I don't expect you to generalize (yet), but as the order of the
derivative on the RHS increases, the output undergoes more drastic changes
from t=0^{-}
to t=0^{+}.)

For the three circuits shown assume the input ei(t)=f(t) is a unit impulse (f(t)=δ(t) and that the system is overdamped so there are no oscillations in the homogeneous response.

Circuit i | |

Circuit ii | |

Circuit iii |

a) Very crudely sketch the impulse response for an overdamped system. Pay particular attention to the response for t near 0 and ∞. Note there are no actual numbers so you can't sketch with any accuracy (but there should be no oscillations in your answer).

b) Answer the following questions. Based on the order of the derivative of the input...

- ...what can you say about the steady state value?
- ...what can you say about the nature of any
discontinuities of the system between t=0
^{-}and t=0^{+}.

a) Note each output looks like the derivative of the one that precedes it (and is the derivative of the step response of the same circuit).

b) As t→∞ the output goes to zero. If the order of the derivative on the right side is
the same as the order of the highest derivative on the left side, then there
is an impulse in the output between t=0^{-} and t=0^{+}.
(Note: I don't expect you to generalize (yet), but as the order of the
derivative on the RHS increases, the output undergoes more drastic changes
from t=0^{-}
to t=0^{+}.)

Consider the circuit below C=0.25F, L=1H, R=5Ω

a) Show that it is described by the differential equation

where f(t)=e_{in}(t), and y(t)=e_{out}(t). A dot above
a function denotes differentiation, 2 dots denotes a second derivative...

b) Show that the unit step response (f(t)=γ(t)) is given by

(remember that the unit step response has zero initial conditions at t=0^{-},
i.e., it is a zero state response).

c) The following MATLAB code plots the step response. Add comments to the code and then turn in the code and the resulting plot. You will be graded on the quality of your comments. The file is at e12AsnStepPlot.m. Note: the MATLAB code "t>=0" will be equal to 0 when t is less than 0 and equal to 1 for t greater than or equal to zero.

a) Sum voltages around loop:

b) Characteristic equation is s^{2}+5s+4=(s+4)(s+1), Characteristic
values, s=-1, s=-4. Assume forms of homogeneous and particular solutions:

Put x_{p}(t) into differential equation

Get complete response for t≥0

Get A and B from initial conditions at t=0^{+}. The voltage across
the capacitor cannot change instantaneously so

and the current through the inductor cannot change instantaneously, so

So

We multiply by the unit step to make the function valid at all times.

c)

t=linspace(-1,5,1000); % Create time vector from -1 to 5 in 1000 steps. plot(t,t>=0,'b','Linewidth',2); % Plot unit step - the value of t>=0 % is 0 for t<0 and 1 otherwise. The plot % will be a thick blue line. hold on % Don't erase first plot % This "plot" statement plots step response multiplied by (t>=0) to make it % equal to 0 for negative values of t. The plot will be a dotted red line. plot(t,(1+exp(-4*t)/3-4/3*exp(-t)).*(t>=0),'r:'); title('Step Response, E12 Assignment'); % Plot title xlabel('Time (s)'); % x-axis label ylabel('Voltage (volts)'); % y-axis label axis([min(t) max(t) -1 2]); % Set axis limits legend('Input','Output'); % Add legend to plot. hold off % Erase plot at next "plot" command.

For the previous problem

a) Find an expression for the unit impulse response.

b) Use Matlab to plot it - turn in well commented Matlab code and the resulting plot (you will be graded partly on the quality of your comments). Use the same time scale as before.

a)

but f(t)δ(t)=f(0)δ(t), so

b)

% Plot the impulse response t=linspace(-1,5,1000); % Create time vector from -1 to 5 in 1000 steps. % This plot statement plots impulse response multiplied by (t>=0) to make it % equal to 0 for negative values of t. The plot will be a dotted red line. plot(t,(exp(-t)-exp(-4*t))*4/3.*(t>=0),'r:'); title('Impulse Response, E12 Assignment'); % Plot title xlabel('Time (s)'); % x-axis label ylabel('Voltage (volts)'); % y-axis label axis([min(t) max(t) -1 2]); % Set axis limits

The circuit shown

with C=0.25F, L=1H, R=5Ω and f(t)=e_{in}(t), and y(t)=e_{out}(t)
is defined by the differential equation

and has the impulse response

The input to the system is f(t)=3γ(t), and the initial conditions are

a) Using the impulse response and the convolution integral, find an expression for the zero state response

b) Find an expression for the zero input response

c) Find the complete response

d) Use MATLAB to plot the zero state, zero input and complete response on one set of axes. Include a legend to differentiate between them. Turn in the plot and the MATLAB code.

a) Find an expression for the zero state response

b) Find an expression for the zero input response

Assume form of solution with zero input (to find 's')

There are two value for s and therefore two components to y_{ZI}.

Get A and B from initial conditions at t=0^{-}.

c) Find the complete response

d) Use MATLAB to plot the zero state, zero input and complete response on one set of axes. Include a legend to differentiate between them. Turn in the plot and the well commented MATLAB code.

t=linspace(0,5,1000); %Create time vector yzs=(3+exp(-4*t)-4*exp(-t)); %Zero state response yzi=(exp(-4*t)-2*exp(-t)); %Zero input response yc=yzs+yzi; %complete response. % The next lines make a plot, label the axes, and add a legend. plot(t,yzs,t,yzi,t,yc); title('Step Response, E12 Assignment'); xlabel('Time (s)'); ylabel('Voltage (volts)'); legend('zero state','zero input','complete');

Given a system with impulse response

Use the convolution integral to find the zero state solution if f(t)=tγ(t) (i.e.. f(t)=t for t≥0). Note: this is a problem you could not have done in E11. Note also that the integration is somewhat long and tedious but not particularly difficult if you recall (i.e., look it up in an integral table) that

That integration was hard – we won’t have too many problems with integrations that are that difficult

A system defined by the differential equation

has the impulse response

a) If the input is a ramp (f(t)=2.5·tγ(t)), and the initial conditions are , find the output without solving any additional differential equations by using the results of previous problem.

b) What is the output if the input is f(t)=-2γ(t), and the initial conditions are ? Do as little additional work as possible.

a) We know know the response to a unit ramp (see a previous problem).
Since the input is now 2.5 times as large, the output will be 2.5=5/2 times
as large. This gives us y_{ZS}. From a previous problem
we also know y_{ZI}
(we've seen these initial conditions before in a previous problem).

b) We know y_{ZS}. This is because we know the response to a
step of height 3. To find the response to a step of height A we multiply
by A/3 (in this case -2/3). But we must
find y_{ZI} from initial conditions

Hint: start with substitution: u=t-λ

Start with the substitution

so

(note the change in sign of limits of
integration when we change the integration variable)

Swap limits of integration and change sign in front of integral

Replace dummy integration variable on right hand side by λ and we are done.