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The questions 270a_hw1.pdf

HW1, MAE 270A. Linear systems I. Fall 2005. UCI

Nasser M. Abbasi

June 21, 2014

Contents

1 Problem 2.2
2 Problem 2.3
3 Problem 2.5
 3.1 part 1
 3.2 part 2
 3.3 part 3

1 Problem 2.2

The impulse response of an ideal lowpass filter is given by

          sin2ω (t − t0)
g (t) = 2ω --2ω-(t −-t-)-
                    0

for all t  , where ω  ,t0   are constants. Is the ideal lowpass filter causal? Is it possible to build the filter in the real world?

Answer

A system is causal when its output measured at time t  depends only on its the input at time t  and possibly input before that.

Equivalently we can say that a system is causal if its impulse response is always zero when observed at any time before the time of applying the impulse itself.

For the above case, assume that the impulse to the lowpass filter was applied at time t0   and let us observe the response of the system at some t < t0   , The output of the system must be zero for all time before t0   for it to be a causal system. (Otherwise we are observing a response before applying the input).

But we can find at least one t < t0   such that g(t) ⁄= 0  , hence not causal

Since it is not causal,it is not possible to build in real world. That is why it is called ideal.

For example, assume t0 = 5  and let t = 4   then we get

          sin2ω (4 − 5)      sin2ω (− 1)
g (4 ) = 2 ω------------ = 2ω ----------- = sin2ω
            2ω (4 − 5)         2ω (− 1)

which is not zero unless ω  was zero. Since we found at least one time instance before the application of the impulse where the output is not zero, this system is not causal.

PIC

2 Problem 2.3

Consider a system whose input u (t)  and y (t)  are related by

                   (
                   |
                   |{  u(t)    t ≤ α
y(t) = Pα (u (t)) =
                   ||(   0      t > α

where α  is a fixed constant. Is this system linear? Time invariant? Causal?

Answer

A system represented by an operator L  is linear if

L [a w (t) + b v (t)] = a L [ w (t)] + b L [ v (t)]

For all constants a,b   and all input u(t),v (t)

Applying the definition shown to a combination of input we get

pict

Hence the system is linear

A system is time invariant if the following is true:

Apply an impulse or any input to the system which can be in some initial state at time t  and observe the output. Assume the output shows after some Δ (t)  time from the time the input is applied. Δ (t)  can be zero or positive.

Now apply the same input again to the same system when it is in the same initial state as before but with some delay δt.  If the same output will result as before and will also appear after the same Δ (t)  from the time the delayed input was applied, then the system is time invariant.

Another way to express this is to say that the by shifting the input by some δt  , the output will remain the same and will be shifted by the same δt  .

This must be true for any δt.

So we need to ask, is

Pα (u(t − τ))  =  y(t − τ)

for any delay τ ?

This system is clearly not time invariant. Suppose we apply a unit step function as the input. The output will then be of height 1 from t = 0  up to α  and will be zero after that. So the output is 1 for a width of α  .

Now if we delay the input by any value, we should also get an output to be of height 1 and of width α  (but shifted by the same delay) if this system to be time variant. But if we delay the input by an amount that happens to be α  , then the output will be zero. hence system is not time invariant

Is it causal?

Assume that the input was applied at the time t0   and let us observe the response of the system at some t < t
     0   , hence the output of the system must be zero for all time before t0   for it to be causal. 2 cases. Assume t0 ≤ α  , then y (t) = u (t)  , hence if u(t)  was zero before t0   then clearly y (t)  will also be zero before t0   since it is the same signal.

Now assume t0 > α  , hence y(t) = 0  by definition.

Hence this is causal system

3 Problem 2.5

Consider a system with input u  and output y  . 3 experiments are performed on the system using the input u ,u ,u
 1  2   3   for t ≥ 0  . In each case the initial state x (0)  time time t = 0  is the same. The corresponding output are denoted by y1,y2,y3   . which is of the following statements are correct if x(0) ⁄= 0?

1.
if u3 = u1 + u2   then y3 = y1 + y2
2.
if u3 = 0.5(u1 + u2)  then y3 = 0.5 (y1 + y2)
3.
if u3 = u1 − u2   then y3 = y1 − y2

which are correct if x (0) = 0?

Answer

Assume system is linear.

Let Yj,zir  means the output for system j  when input is zero

Let Yj,zsr  means the output for system j  when state x (0)  is zero

hence

yj = Yj,zir + Yj,zsr

Since it is the same system, and the same initial state, then Y
  j,zir  is the same for all j  , call it Z  hence

Yj,zir = Z

hence we write

yj = Z + Yj,zir

Where A  is the response of the system when input is zero. i.e. it is the response due to the initial state x (0)  only.

3.1 part 1

when u3 = u1 + u2,  since system is linear then

Y3,zsr = Y1,zsr + Y2,zsr
(1)

Now is y3 = y1 + y2   ?

pict

substitute (1) into (2)   we obtain that Z  = 2Z  . Now when x (0) ⁄= 0  then Z  ⁄= 0  , hence we can divide by Z  and obtain that 1 = 2  which is not correct. Hence (1) is not correct if x(0) ⁄= 0

if x(0) = 0  then Z = 0  , then (2)=(1), hence it is correct in this case.

3.2 part 2

When u3 = 0.5(u1 + u2),  and since system is linear then

Y3,zsr = 0.5(Y1,zsr + Y2,zsr)
(3)

is y3 = 0.5(y1 + y2)  ?

pict

subtract Z  from each side, we see that (4) is the same as (3) Hence (2) is correct if x(0) ⁄= 0  It is also correct if x (0 ) ⁄= 0  since is independent on Z

3.3 part 3

When u3 =  u1 − u2,  and since system is linear then

Y3,zsr = Y1,zsr − Y2,zsr
(5)

is y3 =  y1 − y2   ?

pict

substitute (5) into equation (6)   hence we obtain that Z =  0

If x(0) ⁄= 0  then Z ⁄= 0  hence Hence (3) is not correct if x (0 ) ⁄= 0  but if x (0 ) = 0  then Z  = 0  hence it is correct in that case.