## Contents

### Problem

Consider two systems:

• x(t) → 1 → x(t-7)
• x(t) → 2 → x(2t)

What happens in the following?

a. x(t) → 1 → 2 → ?

b. x(t) → 2 → 1 → ?

### Solution to a.

Define functions x, y, and z as follows:

x → 1 → y → 2 → z

x(t) → 1 → y(t) = x(t-7)

y(t) → 2 → z(t) = y(2t)

z(t) = y(2t)

y(2t) = x((2t)-7) = x(2t-7).

It may be helpful to consider x(t) = t.

If x(t) = t, then y(t) = x(t-7) = t-7.

If y(t) = t-7, then z(t) = y(2t) = (2t) - 7.

Then, we can simply conclude that since x(2t-7) = 2t-7, z(t) = x(2t-7).

### Solution to b.

Next, we move onto b.

Define functions x, y, and z as follows:

x → 2 → y → 1 → z

x(t) → 2 → y(t) = x(2t)

y(t) → 1 → z(t) = y(t-7)

z(t) = y(t-7)

y(t-7) = x(2(t-7)) = x(2t-14).

Again, it may be helpful to consider x(t) = t.

If x(t) = t, then y(t) = x(2t) = 2t.

If y(t) = 2t, then z(t) = y(t-7) = 2(t-7) = 2t-14.

Then, we can simply conclude that since x(2t-14) = 2t-14, z(t) = x(2t-14).

### Note

The answers to solutions in a and b are not the same! A cascaded system cannot be freely swapped around and be expected to behave in the same way.

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale. Dr. Paul Garrett