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raccourcis clavier

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problème 1.

The Root Mean Square (RMS) value of a signal f(t)f(t) that is periodic with period TT is given by the equation 1T0T(f(t))2dt\sqrt{\frac{1}{T} \int_0^T{(f(t))^2dt}} It can be shown that the RMS value of u(t)=Bsinωtu(t) = B \sin{\omega t} is B2\frac{B}{\sqrt{2}}

1.a

Square wave

The square wave function is defined as:

f(t)={1if 0t<T20if T2t<Tf(t) = \begin{cases} 1 & \text{if } 0 \leq t < \frac{T}{2} \\ 0 & \text{if } \frac{T}{2} \leq t < T \end{cases}
import sympy as sp
 
t = sp.symbols('t')
T = 2
 
RMS = sp.sqrt(1/T * sp.integrate(1, (t, 0, T/2)))

RMS = 12\frac{1}{\sqrt{2}}

1.b

Sawtooth wave

A sawtooth wave function is defined as:

f(t)=2AT(tT2)f(t) = \frac{2A}{T}(t - \frac{T}{2})
import sympy as sp
 
t = sp.symbols('t')
T = 1
A = 0.5
 
f_t = 2 * A / T * (t - T/2)
 
RMS = sp.sqrt(1/T * sp.integrate(f_t**2, (t, 0, T)))

RMS = 36\frac{\sqrt{3}}{6}

1.c

sine wave

A general form of the sine wave can be written as

f(t)=Asin(ωt+ϕ)f(t) = A \sin(\omega t + \phi)

Amplitude is 2.3, no phase shift

RMS = 2.32\frac{2.3}{\sqrt{2}}


problème 2.

Find the cutoff frequency of the following low-pass filters.

Cutoff frequency of low-pass filters, the frequency at which the amplitude falls to 120.707\frac{1}{\sqrt{2}} \approx 0.707

2.a

0.05Hz

2.b

approx. 1.1e05 Hz

2.c

approx 1.1Hz