The continuous time signal $x(t)$ is real, periodic with period $T$ and satisfies the Dirichlet conditions.
The Fourier series representation of $x(t) = \sum_{-\infty}^\infty a_n e^{j\left(\frac{2\pi nt}{T}\right)}$ and $x(t)$ satisfies the following: $x\left(t - \frac{T}{2}\right) = -x(t)$.
For any integer $m$, which of the following options is correct?
Solution
The problem concerns a real, periodic continuous-time signal $x(t)$ with period $T$ that satisfies Dirichlet conditions and a specific symmetry property: $x\left(t - \frac{T}{2}\right) = -x(t)$. This property is known as half-wave symmetry. We need to determine the condition for its Fourier series coefficients $a_n$.
Fourier Series Representation
The signal $x(t)$ is given by its complex exponential Fourier series:
$x(t) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t}$where $\omega_0 = \frac{2\pi}{T}$ is the fundamental angular frequency.
Applying the Half-Wave Symmetry
First, let's express $x\left(t - \frac{T}{2}\right)$ using the Fourier series:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 (t - T/2)}$Using $\omega_0 T/2 = \pi$, we get:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t} e^{-j n \pi}$Recall that $e^{-j n \pi} = (-1)^n$. Therefore:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t}$Deriving the Condition on Coefficients
Now, substitute this back into the given symmetry property $x\left(t - \frac{T}{2}\right) = -x(t)$:
$\sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t} = -\sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t}$Combine the terms:
$\sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t} + \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t} = 0$ $\sum_{n=-\infty}^\infty a_n [(-1)^n + 1] e^{j n \omega_0 t} = 0$Due to the linear independence of the complex exponential functions $e^{j n \omega_0 t}$, the coefficients must satisfy:
$a_n [(-1)^n + 1] = 0 \quad \text{for all integers } n$Analyzing Even Index Coefficients
We are interested in the coefficients where the index $n$ is an even integer. Let $n = 2m$, where $m$ is any integer.
For even $n$, $(-1)^n = (-1)^{2m} = 1$. Substituting this into the derived condition:
$a_{2m} [1 + 1] = 0$ $a_{2m} [2] = 0$ $2a_{2m} = 0$This implies that:
$a_{2m} = 0$Note: For odd indices ($n = 2m+1$), the condition becomes $a_{2m+1}[(-1)^{2m+1} + 1] = a_{2m+1}[-1 + 1] = 0$, which simplifies to $0 = 0$. This provides no constraint on the odd coefficients.
Conclusion
The half-wave symmetry property $x\left(t - \frac{T}{2}\right) = -x(t)$ directly leads to the conclusion that all Fourier series coefficients corresponding to even indices must be zero. Thus, $a_{2m} = 0$ for any integer $m$.
Important Questions from Fourier Series
If we use the Fourier transform ϕ(x, y) = \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\) to solve the partial differential equation \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\) in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α and y β . The values of α and β are
When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?
I. Energy
II. Power
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The value of a0 (round off to two decimal places), is
