All questions of Fourier Series in Signals & Systems for Electrical Engineering (EE) Exam

Please explain why we not consider the 2nd option because this is also a irrational number..

Understanding the Signal u(sin(t))
The signal u(sin(t)) is a function that is periodic with a fundamental frequency. To analyze its Fourier series representation, we must consider its symmetry and harmonic content.
Fourier Series Representation
- The Fourier series of a periodic function decomposes it into a sum of sine and cosine functions with different frequencies (harmonics).
- A key aspect of the Fourier series is that it can include both even and odd harmonics, as well as a DC component, depending on the signal's symmetry.
Symmetry Analysis
- The function u(sin(t)) is an odd function, meaning it is symmetric about the origin. This characteristic leads to the following implications:
- It contains only sine terms (which are odd functions) in its Fourier series expansion.
- There is no DC component since a DC term would be represented by a cosine term.
Harmonics Present
- The Fourier series will consist of the following:
- Odd Harmonics of Sine: The periodic nature of sin(t) leads to the presence of odd harmonics (3rd, 5th, etc.) in the expansion.
- DC Component: Since the signal is odd, the average value over one period is zero, leading to no DC term.
Conclusion
- Thus, the Fourier series of u(sin(t)) includes:
- Both odd harmonics of sine.
- Zero DC component.
- Therefore, the correct answer is option 'C': DC and odd harmonics of sine.
This understanding highlights the significance of symmetry in determining the harmonic content of periodic signals in the context of Fourier analysis.

Explanation:
Dirichlet Conditions:
Dirichlet conditions are a set of criteria that a function must satisfy to be expanded in a Fourier series. These conditions are as follows:
1. The function must be single-valued and continuous in the interval for which it is being expanded.
2. The function must have a finite number of maxima and minima in the interval.
3. The function must have a finite number of discontinuities in the interval, which can be of the first kind (jump discontinuities) or the second kind (infinite discontinuities).
4. If the function has discontinuities, then the left and right limits of the function at the discontinuity point must exist and be finite.

f(t) = -f(-t):
The given function f(t) satisfies the odd symmetry property, i.e., f(t) = -f(-t). This means that the function is symmetric about the origin, and its Fourier series will only contain sine terms. This is because the Fourier series of an odd function only contains sine terms, while the Fourier series of an even function only contains cosine terms. Therefore, option A is the correct answer.

Conclusion:
In conclusion, if a function satisfies the odd symmetry property and the Dirichlet conditions, then its Fourier series will only contain sine terms.

Understanding the Problem
The given periodic signal is defined as y[n] = x[n] - x[n + N/2], where x[n] is a periodic signal with period N. Here, N is even, indicating that the signal repeats every N samples.
Fourier Series Coefficients
The Fourier Series (FS) coefficients X[k] represent the frequency components of the periodic signal x[n]. To find the FS coefficients Y[k] of the modified signal y[n], we need to analyze the impact of the subtraction operation in y[n]:
- The term x[n + N/2] represents the signal shifted by half its period.
- Since N is even, x[n + N/2] will have a specific relationship with x[n].
Calculating the Coefficients
Using the properties of Fourier series:
- The FS coefficients of x[n + N/2] can be derived from X[k], specifically:
- X[k + N/2] if k is shifted by N/2.
Thus, for y[n]:
Y[k] = X[k] - X[k + N/2]
Now, we evaluate how this affects the coefficients:
- Since X[k + N/2] is the FS representation of x[n + N/2], and because x[n] is periodic, we can deduce that this will produce a specific pattern in the coefficients.
Identifying the Correct Option
After analysis, we can conclude:
- When analyzing the subtraction, we see that:
- The coefficients will cancel out in certain frequencies depending on the periodic nature.
- The correct expression that represents this relationship in the question is:
Y[k] = (1 - (-1)^k)X[k]
This indicates that Y[k] takes into account the effect of the periodic nature of x[n] and its shifted version. Each term contributes to Y[k] based on the evenness of the index k.
Final Conclusion
Thus, the correct answer is option 'B':
Y[k] = (1 - (-1)^k)X[k].
This option accurately reflects the properties of the periodic signal and its Fourier Series representation.

Understanding the Fourier Series of Odd Functions
When considering the Fourier series expansion of functions, it's important to understand how different types of functions (even, odd) affect the series representation.
Odd Functions
- An odd function is defined by the property f(-x) = -f(x).
- This symmetry implies that the function is symmetric about the origin.
Fourier Series Components
The Fourier series expansion of a periodic function can be expressed as:
- f(x) = a0/2 + Σ [an * cos(nωx) + bn * sin(nωx)]
Where:
- a0 is the average value of the function.
- an are the coefficients for the cosine terms (even functions).
- bn are the coefficients for the sine terms (odd functions).
Why Only Sine Terms for Odd Functions?
- For odd functions, the average value (a0) is zero because the positive and negative areas cancel each other out over one period.
- The cosine terms (an) are eliminated because they represent even functions, which do not satisfy the symmetry of odd functions.
- Thus, only sine terms (bn) remain, representing the odd harmonics of the function.
Conclusion
In conclusion, when performing a Fourier series expansion for an odd function, the result will contain only sine terms. Therefore, the correct answer is:
- a) only sine terms.