An engineer computes a Fourier series coefficient for a periodic load signal. Given peak amplitude (F0) = 11.0000; period (T) = 4.0000 s; harmonic scale (n) = 3.0000, determine the fundamental coefficient (a1).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except a1 is given, so isolate a1 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for a1:

   $$a_{1} = \dfrac{F_0 n}{2}$$

3. Step 3 — List the givens: peak amplitude (F0) = 11.0000, period (T) = 4.0000 s, harmonic scale (n) = 3.0000.

4. Step 4 — Substitute the given values:

   $$a_{1} = \dfrac{F_0 3.0000}{2}$$

5. Step 5 — Evaluate:

   $$a_{1} = 16.5000$$

6. Step 6 — Check: returning a1 = 16.5000 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 33.0000 — kept a factor of two that cancels in the correct rearrangement.
• 8.2500 — dropped that same factor in the other direction.
• 18.1500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

A student expands a square wave in a Fourier series and checks the first coefficient. Given period (T) = 1.0000 s; fundamental coefficient (a1) = 16.4700; harmonic scale (n) = 2.0000, determine the peak amplitude (F0).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except F0 is given, so isolate F0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for F0:

   $$F_{0} = \dfrac{2 a_1}{n}$$

3. Step 3 — List the givens: period (T) = 1.0000 s, fundamental coefficient (a1) = 16.4700, harmonic scale (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$F_{0} = \dfrac{2 a_1}{2.0000}$$

5. Step 5 — Evaluate:

   $$F_{0} = 16.4700$$

6. Step 6 — Check: returning F0 = 16.4700 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 32.9400 — kept a factor of two that cancels in the correct rearrangement.
• 8.2350 — dropped that same factor in the other direction.
• 18.1170 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

The Fourier series of a vibration signal is truncated to its fundamental term. Given peak amplitude (F0) = 5.5000; period (T) = 1.3000 s; fundamental coefficient (a1) = 11.9100, determine the harmonic scale (n).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{2 a_1}{F_0}$$

3. Step 3 — List the givens: peak amplitude (F0) = 5.5000, period (T) = 1.3000 s, fundamental coefficient (a1) = 11.9100.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{2 a_1}{F_0}$$

5. Step 5 — Evaluate:

   $$n = 4.3309$$

6. Step 6 — Check: returning n = 4.3309 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8.6618 — kept a factor of two that cancels in the correct rearrangement.
• 2.1655 — dropped that same factor in the other direction.
• 4.7640 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

An engineer computes a Fourier series coefficient for a periodic load signal. Given peak amplitude (F0) = 15.0000; period (T) = 1.7000 s; harmonic scale (n) = 3.0000, determine the fundamental coefficient (a1).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except a1 is given, so isolate a1 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for a1:

   $$a_{1} = \dfrac{F_0 n}{2}$$

3. Step 3 — List the givens: peak amplitude (F0) = 15.0000, period (T) = 1.7000 s, harmonic scale (n) = 3.0000.

4. Step 4 — Substitute the given values:

   $$a_{1} = \dfrac{F_0 3.0000}{2}$$

5. Step 5 — Evaluate:

   $$a_{1} = 22.5000$$

6. Step 6 — Check: returning a1 = 22.5000 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 45.0000 — kept a factor of two that cancels in the correct rearrangement.
• 11.2500 — dropped that same factor in the other direction.
• 24.7500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

A student expands a square wave in a Fourier series and checks the first coefficient. Given period (T) = 3.2000 s; fundamental coefficient (a1) = 1.1600; harmonic scale (n) = 2.0000, determine the peak amplitude (F0).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except F0 is given, so isolate F0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for F0:

   $$F_{0} = \dfrac{2 a_1}{n}$$

3. Step 3 — List the givens: period (T) = 3.2000 s, fundamental coefficient (a1) = 1.1600, harmonic scale (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$F_{0} = \dfrac{2 a_1}{2.0000}$$

5. Step 5 — Evaluate:

   $$F_{0} = 1.1600$$

6. Step 6 — Check: returning F0 = 1.1600 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.3200 — kept a factor of two that cancels in the correct rearrangement.
• 0.5800 — dropped that same factor in the other direction.
• 1.2760 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

The Fourier series of a vibration signal is truncated to its fundamental term. Given peak amplitude (F0) = 9.5000; period (T) = 1.8000 s; fundamental coefficient (a1) = 17.7900, determine the harmonic scale (n).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{2 a_1}{F_0}$$

3. Step 3 — List the givens: peak amplitude (F0) = 9.5000, period (T) = 1.8000 s, fundamental coefficient (a1) = 17.7900.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{2 a_1}{F_0}$$

5. Step 5 — Evaluate:

   $$n = 3.7453$$

6. Step 6 — Check: returning n = 3.7453 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 7.4905 — kept a factor of two that cancels in the correct rearrangement.
• 1.8726 — dropped that same factor in the other direction.
• 4.1198 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

An engineer computes a Fourier series coefficient for a periodic load signal. Given peak amplitude (F0) = 19.5000; period (T) = 2.9000 s; harmonic scale (n) = 1.0000, determine the fundamental coefficient (a1).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except a1 is given, so isolate a1 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for a1:

   $$a_{1} = \dfrac{F_0 n}{2}$$

3. Step 3 — List the givens: peak amplitude (F0) = 19.5000, period (T) = 2.9000 s, harmonic scale (n) = 1.0000.

4. Step 4 — Substitute the given values:

   $$a_{1} = \dfrac{F_0 1.0000}{2}$$

5. Step 5 — Evaluate:

   $$a_{1} = 9.7500$$

6. Step 6 — Check: returning a1 = 9.7500 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 19.5000 — kept a factor of two that cancels in the correct rearrangement.
• 4.8750 — dropped that same factor in the other direction.
• 10.7250 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

A student expands a square wave in a Fourier series and checks the first coefficient. Given period (T) = 1.2000 s; fundamental coefficient (a1) = 11.6400; harmonic scale (n) = 2.0000, determine the peak amplitude (F0).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except F0 is given, so isolate F0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for F0:

   $$F_{0} = \dfrac{2 a_1}{n}$$

3. Step 3 — List the givens: period (T) = 1.2000 s, fundamental coefficient (a1) = 11.6400, harmonic scale (n) = 2.0000.

4. Step 4 — Substitute the given values:

   $$F_{0} = \dfrac{2 a_1}{2.0000}$$

5. Step 5 — Evaluate:

   $$F_{0} = 11.6400$$

6. Step 6 — Check: returning F0 = 11.6400 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 23.2800 — kept a factor of two that cancels in the correct rearrangement.
• 5.8200 — dropped that same factor in the other direction.
• 12.8040 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

The Fourier series of a vibration signal is truncated to its fundamental term. Given peak amplitude (F0) = 7.5000; period (T) = 2.0000 s; fundamental coefficient (a1) = 3.6800, determine the harmonic scale (n).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{2 a_1}{F_0}$$

3. Step 3 — List the givens: peak amplitude (F0) = 7.5000, period (T) = 2.0000 s, fundamental coefficient (a1) = 3.6800.

4. Step 4 — Substitute the given values:

   $$n = \dfrac{2 a_1}{F_0}$$

5. Step 5 — Evaluate:

   $$n = 0.9813$$

6. Step 6 — Check: returning n = 0.9813 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.9627 — kept a factor of two that cancels in the correct rearrangement.
• 0.4907 — dropped that same factor in the other direction.
• 1.0795 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series

An engineer computes a Fourier series coefficient for a periodic load signal. Given peak amplitude (F0) = 6.0000; period (T) = 2.8000 s; harmonic scale (n) = 1.0000, determine the fundamental coefficient (a1).

Start with the thinking

• The governing relation printed in this handbook section is Fourier series — fundamental coefficient.
• Everything except a1 is given, so isolate a1 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Fourier series fundamental coefficient a1 is scaled from the periodic signal's peak amplitude.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

2. Step 2 — Rearrange symbolically for a1:

   $$a_{1} = \dfrac{F_0 n}{2}$$

3. Step 3 — List the givens: peak amplitude (F0) = 6.0000, period (T) = 2.8000 s, harmonic scale (n) = 1.0000.

4. Step 4 — Substitute the given values:

   $$a_{1} = \dfrac{F_0 1.0000}{2}$$

5. Step 5 — Evaluate:

   $$a_{1} = 3.0000$$

6. Step 6 — Check: returning a1 = 3.0000 to

   $$a_1 = \dfrac{2}{T} \int_0^T f(t) \cos\!\left(\dfrac{2\pi t}{T}\right) dt$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 6.0000 — kept a factor of two that cancels in the correct rearrangement.
• 1.5000 — dropped that same factor in the other direction.
• 3.3000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fourier Series