* The Probability density function (PA) of a stgnal is ginen by ke PCx) 2 Ix1 <4 othermse a) what the step size 19) f tvere are 4 quantitatron leuls. b) fmd the voolue f Conshont K solution :) Sten 1 of 3) The sign is inside the absolute value its sign must be positive regardless of the sign inside the absolute, why did the sign come out outside the absolute value and why is it distributed once positive and again negative? VH = keee Ke = K ke 4 n= 4 e4 Step size VH - VL %3D K - K e 4 in = 0.06135 Step 3 of 3:) k. 4 | ke*dz + う Ke - 4

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* The Probability density functron (PA) of a stgnal
is ginen by
ke
Pcx) =
otheruse
a) what the step size 19) f trere are 4
quantitatron leuls.
) hmd the vodue f Conflont K
solution :)
Sten 1 of 3-
The sign is inside the absolute value
its sign must be positive regardless of
the sign inside the absolute, why did the
sign come out outside the absolute value
and why is it distributed once positive and
again negative?
VH = kege Ke
= K
ke 4
n= 4
e4
Step size
VH - VL
K - K
in
= 0.0613s
a 4
Step 3 of 3:)
k.
4
( ke*dz +
う
Ke
- 4
Transcribed Image Text:* The Probability density functron (PA) of a stgnal is ginen by ke Pcx) = otheruse a) what the step size 19) f trere are 4 quantitatron leuls. ) hmd the vodue f Conflont K solution :) Sten 1 of 3- The sign is inside the absolute value its sign must be positive regardless of the sign inside the absolute, why did the sign come out outside the absolute value and why is it distributed once positive and again negative? VH = kege Ke = K ke 4 n= 4 e4 Step size VH - VL K - K in = 0.0613s a 4 Step 3 of 3:) k. 4 ( ke*dz + う Ke - 4
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