This question was previously asked in

VSSC (ISRO) Technician B: Electronic Mechanic Previous Year Paper (Held on 10 Dec 2017)

Option 2 : 0.02 msec

Subject Test 1: Electrical Basics

3053

20 Questions
80 Marks
30 Mins

**Concept:**

For envelope detector, the time constant must satisfy the following relation:

\(RC \le \frac{1}{{{\omega _m}}}\left[ {\frac{{\sqrt {1 - {μ^2}} }}{μ}\;} \right]\) ---(1)

For the good performance of the envelope detector:

\(\frac{1}{\omega_c}<<RC<<\frac{1}{\omega_m}\)

**Application:**

Given:

f_{m} = 10 kHz

modulation index (μ) = 0.6

From (1), we get

\(RC \le \frac{1}{{{2\pi \times 10}}}\left[ {\frac{{\sqrt {1 - {0.6^2}} }}{0.6}\;} \right]\)

**= 0.02 msec**

__ Derivation of RC time constant formula for the envelope detector:__ )

For the output of the envelope detector to closely follow the modulating signal, the slope of vo is always greater than that of the envelope of AM signal input.

The output vo is the voltage across the capacitor which is exponentially decaying

\(\begin{array}{l} {v_0}\left( t \right) = {v_i}{e^{ - \frac{t}{{RC}}}}\\ = {V_i}\left[ {1 - \frac{t}{{RC}} + \frac{1}{{2!{{\left( {\frac{t}{{RC}}} \right)}^2}}} + \ldots } \right] \end{array}\)

Neglecting higher power terms as RC>>t

\(= {V_i}\left[ {1 - \frac{t}{{RC}}} \right]\)

For successful detection, the rate of discharge of the capacitor i.e. magnitude of the slope of v0(t) must be more than that of the envelope

The envelope of the AM signal is \({V_i}\left( t \right) = {A_c}\left[ {1 + {A_m}\cos {\omega _m}t} \right]\)

The slope of the message signal

\( \frac{{d{V_i}\left( t \right)}}{{dt}} = - {A_c}{A_m}\sin {\omega _m}t\)

The slope of output voltage

\(\frac{{d{v_0}}}{{dt}} = \frac{{{-V_i}}}{{Rc}}\)

For DETECTION slope of capacitor voltage should be always more than the envelope of AM signal

\(\begin{array}{l} \frac{{{V_i}}}{{RC}} \ge - {A_C}{A_m}\sin {\omega _m}t\\ \frac{{{A_c}}}{{RC}}\left[ {1 + {A_m}\cos {\omega _m}t} \right] \le {A_c}{A_m}\sin {\omega _m}t\\ RC \le \frac{{1 + {A_m}\cos {\omega _m}t}}{{{A_m}{\omega _m}\sin {\omega _m}t}} \end{array}\)

RC must be less than the minimum of R.H.S.

\(RC \le \frac{1}{{{\omega _m}}}\left[ {\frac{{\sqrt {1 - {μ^2}} }}{μ}\;} \right]\)