E12 Assignment 01

Problems: 
1. E11 Review - Diff Eq for First Order Circuit

For the circuit shown,

show that if R1=R2=C=1 that

Solution: 

 

2. E11 Review - First Order Circuit with Polynomial Input

The circuit shown (with R1=R2=C=1) is characterized by the differential equation shown

                         

Assume ein(t)=0 for t≤0.  If eout(0-)=0 and ein(t) = 6·t2  + 2   (for t≥0):

  1. Find the form of the homogeneous response.
  2. Find the form of the particular response.
  3. Find the unknown coefficients in the particular response.
  4. Find eout(t).
Solution: 

a)

b) Particular solution is same form as input and its derivatives

c)

d)

3. E11 Review - 1st Order Circuit, exponential input

The circuit shown (with R1=R2=C=1) is characterized by the differential equation shown

                         

Assume ein(t)=0 for t≤0.  If eout(0-)=0 and ein(t) = 8·e2t    (for t≥0):

  1. Find the form of the homogeneous response.
  2. Find the form of the particular response.
  3. Find the unknown coefficients in the particular response.
  4. Find eout(t).
Solution: 

a)

b) Particular solution is same form as input and its derivatives 

c)

d)

4. E11 Review - 1st order circuit w/ exponential input with time constant the same as the circuit

The circuit shown (with R1=R2=C=1) is characterized by the differential equation shown

                         

Assume ein(t)=0 for t≤0.  If eout(0-)=0 and ein(t) = 8·e-2t    (for t≥0):

  1. Find the form of the homogeneous response.
  2. Find the form of the particular response.
  3. Find the unknown coefficients in the particular response.
  4. Find eout(t).
Solution: 

a)

b) Particular solution is same form as input and its derivatives, but in this case the input is the same form as the homogeneous response, so we must multiply it by t. 

c)

d)

5. E11 Review (1st order step)

Consider the differential equation

a) The particular solution is the same form as the input and represents the solution as t→∞.  In this case we have a constant input, so yp(t)=K.  What is the value of K?

b) For first order constant coefficient differential equations we assume a homogeneous solution (RHS of equation = 0) of yh(t)=Aest.  What value of 's' satisfies the differential equation?

c) The complete solution is given by y(t)=yp(t)+yh(t).  If we have y(0+)=1, determine the value of the constant A in yh(t).
Note: y(0+) is the value of y(t) at t=0+, i.e., an infinitesimally small time after t=0.

d) What is y(t)?

Solution: 

a) Put yp(t)=K into differential equation and use the fact that it's derivative is zero.

b) Put yh(t)=Aest into differential equation, set the RHS to zero, and solve for s

c)

d)

6. E11 Review (2nd Order Step)

Consider the differential equation

 

a) The particular solution is the same form as the input and represents the solution as t→∞.  In this case we have a constant input, so yp(t)=K.  What is the value of K?

b) For constant coefficient differential equations we generally assume a homogeneous solution (RHS of equation = 0) of yh(t)=Aest.  What value(s) of s satisfy the homogeneous differential equation?  What is the form of the homogeneous response?

c) The complete solution is given by y(t)=yp(t)+yh(t).  Determine the value of the constant(s) A in yh(t) if

d) What is y(t)?

Solution: 

a) Put yp(t)=K into differential equation and use the fact that it's derivative is zero.

 

b) Put yh(t)=Aest into differential equation, set the RHS to zero, and solve for s

 

c)

d)

7. E11 Review (Step - Critical Damping)

Consider the differential equation

 

a) The particular solution is the same form as the input and represents the solution as t→∞.  In this case we have a constant input, so yp(t)=K.  What is the value of K?

b) If we assume a homogeneous solution (RHS of equation = 0) of yh(t)=Aest.  What value(s) of s satisfy the homogeneous differential equation?  What is the form of the homogeneous response?

c) The complete solution is given by y(t)=yp(t)+yh(t).  Determine the value of the constant(s) A in yh(t) if

d) What is y(t)?

Solution: 

a) Put yp(t)=K into differential equation and use the fact that it's derivative is zero.

b) Put yh(t)=Aest into differential equation, set the RHS to zero, and solve for s.  There is only one value of "s" so system is critically damped.  Use appropriate form of homogenous equation (i.e., time multiplied exponential).

 

c)

d)

8. E11 Review (Short Answer)

When solving a differential equation such as :

where y(t) is the unknown function and f(t) is a known input:

  1. What does the particular response represent?
  2. Given an input function, how is the form of the particular response chosen?
  3. What does the homogeneous response represent?
  4. How is the homogeneous response chosen?
  5. For this third order example, what are possible forms of the homogeneous response? 
    For example, a second order system has three possible forms:
    1. If there are two distinct, real roots (s1, s2) the solution is of the form: 
                            
    2. If there is a single real root, repeated once (s1) the solution is of the form:
                            
    3. If there is a pair of complex conjugate roots (s1,2=-σ±jω) the solution is of the form:
                            
Solution: 
  1. The particular response is the response due to the input as t→∞ (i.e., after all the transients have died out).
  2. The form of the particular response is based on the form of the input function, f(t).  It is essentially a guess (the "uniqueness theorem" states that only a single solution to such a problem exists, so if we guess one that works, it must be correct).
  3. The homogeneous response is the characteristic response of the system when there are no inputs.  It has no real physical meaning, but is mathematically useful.
  4. The homogeneous response is also a guess (that happens to work).  We assume yh(t)=Aest, and we set f(t)=0.  This yields a third order polynomial in s.
  5. There are several possible combinations: