SME
Signals & Systems
Continuous and discrete signals
Use this for “Find the period, angular frequency and value at t = 5 ms of x(t) = 3 cos(100πt + 30°).”
Chart
x(5) = −1.5
This is an example. Type your own values to start a new problem.
T (ms)
=
1
÷
f (Hz)
ω (rad/s)
=
2π
×
f (Hz)
x (V)
=
A (V)
×
cos(360°
×
f (Hz)
×
t (ms)
+
φ
°)
x(t) = A cos(ωt + φ): ω is the number before t, and f = ω ÷ 2π. With φ in degrees, ωt in degrees is 360° × f × t.
Formulas
T
=
1
f
0.02
=
1
50
ω
=
2π
×
f
314.1593 = 2π × 50
x
=
A
×
cos(360°
×
f
×
t
+
φ
°)
−1.5 = 3 × cos(360° × 50 × 0.005 + 30°)
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
The signal is x(t) = A cos(ωt + φ) with A the peak; a sine is a cosine shifted −90°.
•
The phase φ is in degrees, so the steps turn ωt into degrees too: ωt = 360° × f × t.
•
One period T holds one full cycle: T = 1 ÷ f and ω = 2πf in rad/s.
Step-by-step
WE KNOW
A = 3 V, f = 50 Hz, φ = +30°, t = 5 ms
FIND
period (T), angular frequency (ω), value at t (x)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s).
t = 5 ms = 0.005 s (1 s = 1000 ms)
Step 2 · Find period (T)
Use T = 1 ÷ f
One
cycle
takes
1
f
seconds.
T
=
1
50
T
=
0.02
s
Step 3 · Find angular frequency (ω)
Use ω = 2π × f
Each cycle turns the angle by 2π radians.
ω
=
2π
×
50
ω
=
314.1593
rad/s
Step 4 · Find value at t (x)
Use x = A × cos(360° × f × t + φ°)
ωt
in
degrees
is
360°
×
f
×
t
;
add
the
phase,
take
the
cosine
and
scale
by
A
.
x
=
3
×
cos(360°
×
50
×
0.005
+
30°)
x
=
3
×
cos(18,000°
×
0.005
+
30°)
x
=
3
×
cos(90°
+
30°)
x
=
3
×
cos(120°)
x
=
3
×
(−0.5)
x
=
−1.5
V
Step 5 · Convert the answers
Change each answer back to the unit you chose.
T = 0.02 s = 20 ms (1 s = 1000 ms)
Check
Put the numbers back in. Both sides should match.
0.02
=
1
50
✓
314.1593 = 2π × 50 ✓
−1.5 = 3 × cos(360° × 50 × 0.005 + 30°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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