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Vitek1552 [10]
3 years ago
11

A phase modulator produces a maximum phase shift of 45 degrees. The modulation frequency range is 300 to 4000 Hz. What is the ma

ximum frequency deviation possible
Mathematics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

Maximum frequency deviation is 2616.29509 Hz

Maximum frequency deviation=(4000-3000)cos45

Max frequency deviation=2616.29509 Hz

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Step-by-step explanation:

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Which of the following has a solution set of {x | x = 0}?
notka56 [123]

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(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a)  x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1  -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also,   x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

8 0
3 years ago
Read 2 more answers
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