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CT Linear Systems and Differential Equations

Module by: Michael Haag, Richard Baraniuk

Summary: The module discusses how to represent linear, time-invariant systems.

Continuous-Time Linear Systems

Physically realizable, linear time-invariant systems can be described by a set of linear differential equations (LDEs):
ctlin.png
Figure 1: Graphical description of a basic linear time-invariant system with an input, ft f t and an output, yt y t .
dndtnyt+ a n − 1 dn-1dtn-1yt+…+ a 1 ddtyt+ a 0 yt= b m dmdtmft+…+ b 1 ddtft+ b 0 ft n t y t a n − 1 n 1 t y t … a 1 t y t a 0 y t b m m t f t … b 1 t f t b 0 f t Equivalently,
∑i=0n a i didtiyt=∑i=0m b i didtift i n 0 a i i t y t i m 0 b i i t f t (1)
with a n =1 a n 1 .
It is easy to show that these equations define a system that is linear and time invariant. A natural question to ask, then, is how to find the system's output response yt y t to an input ft f t . Recall that such a solution can be written as yt= y i t+ y s t y t y i t y s t We refer to y i t y i t as the zero-input response -- the homogeneous solution due only to the initial conditions of the system. We refer to y s t y s t as the zero-state response -- the particular solution in response to the input ft f t . We now discuss how to solve for each of these components of the system's response.

Finding the Zero-Input Response

The zero-input response, y i t y i t , is the system response due to initial conditions only.
Example 1: Zero-Input Response 
Close the switch in the circuit pictured in Figure 2 at time t=0 and then leave everything else alone. The voltage response is shown in Figure 3.
output.png
Figure 2
response.png
Figure 3
Example 2: Zero-Input Response 
Imagine a mass attached to a spring, as shown in Figure 4. When you pull the mass up and let it go, you have an example of a zero-input response. A plot of this response is shown in Figure 5.
mass.png
Figure 4
sinresponse.png
Figure 5
There is no input, so we solve for y 0 t y 0 t such that
∀ a n , a n =1:∑i=0n a i didti y 0 t=0 a n a n 1 i n 0 a i i t y 0 t 0 (2)
If DD is the derivative operator, we can write the previous equation as:
Dn+ a n − 1 Dn-1+…+ a 0 y 0 t=0 D n a n − 1 D n 1 … a 0 y 0 t 0 (3)
Since we need the weighted sum of a bunch of y 0 t y 0 t 's derivatives to be 00 for all tt, then y 0 tddt y 0 td2dt2 y 0 t… y 0 t t y 0 t 2 t y 0 t … must all be of the same form.
Only the exponential, ⅇst s t where s∈ℂ s , has this property (see your Differential Equation's textbook for details). So we must assume that,
∀c,c≠0: y 0 t=cⅇst c c 0 y 0 t c s t (4)
for some cc and ss.
Since ddt y 0 t=csⅇst t y 0 t c s s t , d2dt2 y 0 t=cs2ⅇst 2 t y 0 t c s 2 s t , … we have Dn+ a n − 1 Dn-1+…+ a 0 y 0 t=0 D n a n − 1 D n 1 … a 0 y 0 t 0
csn+ a n − 1 sn-1+…+ a 1 s+ a 0 ⅇst=0 c s n a n − 1 s n 1 … a 1 s a 0 s t 0 (5)
Equation 5 holds for all tt only when
sn+ a n − 1 sn-1+…+ a 1 s+ a 0 =0 s n a n − 1 s n 1 … a 1 s a 0 0 (6)
Where this equation is referred to as the characteristic equation of the system. The possible values of ss are the roots of this polynomial s 1 s 2 … s n s 1 s 2 … s n s- s 1 s- s 2 s- s 3 …s- s n =0 s s 1 s s 2 s s 3 … s s n 0 i.e. possible solutions are c 1 ⅇ s 1 t c 1 s 1 t , c 2 ⅇ s 2 t c 2 s 2 t , ……, c n ⅇ s n t c n s n t . Since the system is linear, the general solution if of the form:
y 0 t= c 1 ⅇ s 1 t+ c 2 ⅇ s 2 t+…+ c n ⅇ s n t y 0 t c 1 s 1 t c 2 s 2 t … c n s n t (7)
Then, solve for the c 1 … c n c 1 … c n using the initial conditions.
Example 3 
See Lathi p.108 for a good example!
We generally assume that the IC's of a system are zero, which implies y i t=0 y i t 0 . However, the method of solving for y i t y i t will prove useful later on.

Finding the Zero-State Response

Solving a linear differential equation
∑i=0n a i didtiyt=∑i=0m b i didtift i n 0 a i i t y t i m 0 b i i t f t (8)
given a specific input ft f t is a difficult task in general. More importantly, the method depends entirely on the nature of ft f t ; if we change the input signal, we must completely re-solve the system of equations to find the system response.
Convolution helps to bypass these difficulties. In section 2, we explain how convolution helps to determine the system's output, given only the input ft f t and the system's impulse response, ht h t .
Before deriving the convolution procedure, we show that a system's impulse response is easily derived from its linear, differential equation (LDE). We will show the derivation for the LDE below, where m<n m n :
dndtnyt+ a n − 1 dn-1dtn-1yt+…+ a 1 ddtyt+ a 0 yt= b m dmdtmft+…+ b 1 ddtft+ b 0 ft n t y t a n − 1 n 1 t y t … a 1 t y t a 0 y t b m m t f t … b 1 t f t b 0 f t (9)
We can rewrite Equation 9 as
Q D yt= P D ft Q D y t P D f t (10)
where Q D · Q D · is an operator that maps yt y t to the left hand side of Equation 9
Q D yt=dndtnyt+ a n − 1 dn-1dtn-1yt+…+ a 1 ddtyt+ a 0 yt Q D y t n t y t a n − 1 n 1 t y t … a 1 t y t a 0 y t (11)
and P D · P D · maps ft f t to the right hand side of Equation 9. Lathi shows (in Appendix 2.1) that the impulse response of the system described by Equation 9 is given by:
ht= b n δt+ P D y n tμt h t b n δ t P D y n t μ t (12)
where for m<n m n we have b n =0 b n 0 . Also, y n y n equals the zero input response with initial conditions. y n − 1 0=1 y n − 2 0=1…y0=0 y n − 1 0 1 y n − 2 0 1 … y 0 0

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