A continuous time periodic signal $x(t)$ is
$x(t) = 1 + 2 \cos 2\pi t + 2 \cos 4\pi t + 2 \cos 6\pi t$
If $T$ is the period of $x(t)$, then $\frac{1}{T} \int_{0}^{T}|x(t)|^2 dt =$ ___________ (round off to the nearest integer).
Signal Components and Period Determination
The given continuous-time signal is: $x(t) = 1 + 2 \cos(2\pi t) + 2 \cos(4\pi t) + 2 \cos(6\pi t)$
This signal consists of a DC component and three cosine terms.
- DC component: $1$
- First cosine term: $2 \cos(2\pi t)$ (Frequency $f_1 = 1$ Hz, Period $T_1 = 1$ s)
- Second cosine term: $2 \cos(4\pi t)$ (Frequency $f_2 = 2$ Hz, Period $T_2 = 1/2$ s)
- Third cosine term: $2 \cos(6\pi t)$ (Frequency $f_3 = 3$ Hz, Period $T_3 = 1/3$ s)
The fundamental period $T$ of the sum of periodic signals is the least common multiple (LCM) of their individual periods.
Therefore, $T = \text{LCM}(T_1, T_2, T_3) = \text{LCM}(1, 1/2, 1/3) = 1$ second.
Average Power Calculation
The average power $P_{avg}$ is defined as:
$P_{avg} = \frac{1}{T} \int_{0}^{T}|x(t)|^2 dt$Because the cosine terms are orthogonal over the fundamental period $T$, the average power of the sum is the sum of the average powers of its components.
- The average power of the DC component $A$ is $A^2$. For $A=1$, the power is $1^2 = 1$.
- The average power of a cosine term $A \cos(\omega t)$ is $A^2/2$.
- For $2 \cos(2\pi t)$, the average power is $\frac{2^2}{2} = 2$.
- For $2 \cos(4\pi t)$, the average power is $\frac{2^2}{2} = 2$.
- For $2 \cos(6\pi t)$, the average power is $\frac{2^2}{2} = 2$.
Summing the average powers:
$P_{avg} = 1 + 2 + 2 + 2 = 7$Final Result
The calculated average power is $7$.
Rounding $7$ to the nearest integer results in $7$.
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
The trigonometric Fourier series of a periodic time function can have
The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has
The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)
The value of a0 (round off to two decimal places), is
