Fourier Transform

Mathematics · FE Reference Handbook section

Mathematics

6 formulas

10 exam-style examples

~57 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The Fourier transform pair, one form of which is
can be used to characterize a broad class of signal models in terms of their frequency or spectral content. Some useful transform
Some mathematical liberties are required to obtain the second and fourth form. Other Fourier transforms are derivable from the

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

First-order transfer function: Laplace transform and step response — Fourier Transform

A sensor has the transfer function G(s) = 5/(2.3s + 1), where K is the gain. A step input of magnitude 7 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 0.5 s.

Given

$K = 5$
$\tau = 2.3 s$
$Step magnitude = 7$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [5/(2.3s + 1)]·(7/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 5 \times 7 = 35.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(0.5) = 35.00(1 - e^{-0.5/2.3}) = 6.838$

Answer:

y(\infty) = 35.00; y(0.5 s) = 6.838

Why the other options are there

28.162 (decay instead of rise)
0.714 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 2

First-order transfer function: Laplace transform and step response — Fourier Transform (2)

A sensor has the transfer function G(s) = 9/(2.0s + 1), where K is the gain. A step input of magnitude 4 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 1.0 s.

Given

$K = 9$
$\tau = 2.0 s$
$Step magnitude = 4$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [9/(2.0s + 1)]·(4/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 9 \times 4 = 36.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(1.0) = 36.00(1 - e^{-1.0/2.0}) = 14.165$

Answer:

y(\infty) = 36.00; y(1.0 s) = 14.165

Why the other options are there

21.835 (decay instead of rise)
2.250 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 3

First-order transfer function: Laplace transform and step response — Fourier Transform (3)

A sensor has the transfer function G(s) = 3/(2.8s + 1), where K is the gain. A step input of magnitude 3 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 2.5 s.

Given

$K = 3$
$\tau = 2.8 s$
$Step magnitude = 3$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [3/(2.8s + 1)]·(3/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 3 \times 3 = 9.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(2.5) = 9.00(1 - e^{-2.5/2.8}) = 5.315$

Answer:

y(\infty) = 9.00; y(2.5 s) = 5.315

Why the other options are there

3.685 (decay instead of rise)
1.000 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 4

First-order transfer function: Laplace transform and step response — Fourier Transform (4)

A sensor has the transfer function G(s) = 6/(1.4s + 1), where K is the gain. A step input of magnitude 10 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 3.0 s.

Given

$K = 6$
$\tau = 1.4 s$
$Step magnitude = 10$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [6/(1.4s + 1)]·(10/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 6 \times 10 = 60.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(3.0) = 60.00(1 - e^{-3.0/1.4}) = 52.961$

Answer:

y(\infty) = 60.00; y(3.0 s) = 52.961

Why the other options are there

7.039 (decay instead of rise)
0.600 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 5

First-order transfer function: Laplace transform and step response — Fourier Transform (5)

A sensor has the transfer function G(s) = 8/(1.4s + 1), where K is the gain. A step input of magnitude 6 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 2.5 s.

Given

$K = 8$
$\tau = 1.4 s$
$Step magnitude = 6$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [8/(1.4s + 1)]·(6/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 8 \times 6 = 48.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(2.5) = 48.00(1 - e^{-2.5/1.4}) = 39.951$

Answer:

y(\infty) = 48.00; y(2.5 s) = 39.951

Why the other options are there

8.049 (decay instead of rise)
1.333 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 6

First-order transfer function: Laplace transform and step response — Fourier Transform (6)

A sensor has the transfer function G(s) = 4/(1.5s + 1), where K is the gain. A step input of magnitude 2 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 3.0 s.

Given

$K = 4$
$\tau = 1.5 s$
$Step magnitude = 2$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [4/(1.5s + 1)]·(2/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 4 \times 2 = 8.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(3.0) = 8.00(1 - e^{-3.0/1.5}) = 6.917$

Answer:

y(\infty) = 8.00; y(3.0 s) = 6.917

Why the other options are there

1.083 (decay instead of rise)
2.000 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 7

First-order transfer function: Laplace transform and step response — Fourier Transform (7)

A sensor has the transfer function G(s) = 2/(0.7s + 1), where K is the gain. A step input of magnitude 6 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 1.5 s.

Given

$K = 2$
$\tau = 0.7 s$
$Step magnitude = 6$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [2/(0.7s + 1)]·(6/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 2 \times 6 = 12.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(1.5) = 12.00(1 - e^{-1.5/0.7}) = 10.592$

Answer:

y(\infty) = 12.00; y(1.5 s) = 10.592

Why the other options are there

1.408 (decay instead of rise)
0.333 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 8

First-order transfer function: Laplace transform and step response — Fourier Transform (8)

A sensor has the transfer function G(s) = 5/(0.2s + 1), where K is the gain. A step input of magnitude 9 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 1.5 s.

Given

$K = 5$
$\tau = 0.2 s$
$Step magnitude = 9$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [5/(0.2s + 1)]·(9/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 5 \times 9 = 45.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(1.5) = 45.00(1 - e^{-1.5/0.2}) = 44.975$

Answer:

y(\infty) = 45.00; y(1.5 s) = 44.975

Why the other options are there

0.025 (decay instead of rise)
0.556 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 9

First-order transfer function: Laplace transform and step response — Fourier Transform (9)

A sensor has the transfer function G(s) = 8/(1.2s + 1), where K is the gain. A step input of magnitude 10 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 3.0 s.

Given

$K = 8$
$\tau = 1.2 s$
$Step magnitude = 10$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [8/(1.2s + 1)]·(10/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 8 \times 10 = 80.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(3.0) = 80.00(1 - e^{-3.0/1.2}) = 73.433$

Answer:

y(\infty) = 80.00; y(3.0 s) = 73.433

Why the other options are there

6.567 (decay instead of rise)
0.800 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform

Example 10

First-order transfer function: Laplace transform and step response — Fourier Transform (10)

A sensor has the transfer function G(s) = 5/(1.8s + 1), where K is the gain. A step input of magnitude 10 is applied at t = 0. Find the Laplace-domain output, the final value, and the response at t = 3.5 s.

Given

$K = 5$
$\tau = 1.8 s$
$Step magnitude = 10$

Find

Y(s), final value, and y(t) at the stated time

Start with the thinking

The Laplace transform of a step of magnitude A is A/s.
The final-value theorem gives the steady state without inverting the transform.

Step-by-step solution

Formula — Y(s) = G(s)·U(s) = [5/(1.8s + 1)]·(10/s)
Final-value theorem — y(∞) = lim(s→0) s·Y(s)
Substituting
$y(\infty) = 5 \times 10 = 50.00$
Inverse transform
$y(t) = K A (1 - e^{-t/\tau})$
Substituting
$y(3.5) = 50.00(1 - e^{-3.5/1.8}) = 42.847$

Answer:

y(\infty) = 50.00; y(3.5 s) = 42.847

Why the other options are there

7.153 (decay instead of rise)
0.500 (divided by the input)

Reference: FE Reference Handbook — Mathematics → Fourier Transform