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# Analog|Definition & Meaning

## Definition

The word **analog** refers to a **process** or **equipment** that **represents** information using **continually** changing **physical** values. **Analog technology** uses physical **quantities** like **voltage** to measure, store, or record an **eternally** changeable **amount** of **information.**

Therefore, it refers to something **tangible** that is **constantly** changing.

## Description of Analog

**The **term analog is used mainly for **signals.** The **analog signal** represents the **information** in continuous form. The **devices** used for **analog signals** are called **analog devices.** In **everyday** situations, we use them **everywhere.**

We use the **wave-based** diagram below to **illustrate** the idea of an **analog** signal:

**Figure** 1 shows the **analog** wave of **cos(x)** where the **magnitude** of this **analog** wave is 2.

Let us consider an example of an **analog telephone.** The human voice is **transmitted** across **analog telephone** lines as **electrical impulses.** The **microphone** on your phone **transforms** the sound waves into analog electrical **waves** whenever you talk into the device.

These waves go to their destination via the **telephone** line, which acts as their medium of propagation. The **speaker** in the receiving phone’s handset completes the process of converting the electrical impulses back into **sound** waves.

**Information** that is **analog** in its original form (such as audio and images) can display a **continuous** range of **shifts** in terms of intensity (such as loudness or luminance) and **frequency.** The analog electrical network converts the **informational** fluctuations in the native **information** stream into amplitude and **frequency** shifts in the carrier signal.

In other words, in order to produce an analog of the initial information stream, the carrier **signal** must be **modulated,** which means that its **frequency** is changed.

**Amplitude modulation** is a **technique** that allows for changes in the amplitude of **electromagnetic sinusoidal** (waveforms), often known as sine waves, while **maintaining** the same **frequency** (AM).

Alternately, a technique called **frequency modulation** can be used to alter the frequencies of the **sine** wave while keeping the **amplitude** the same (FM). In addition, it is possible to modify both the **frequency** and the amplitude at the **same** time.

A **feature** of the **medium** is used by an **analog** signal in order to communicate the **information** being sent by the **signal.**

## Visual Explanation of Analog

Figure 2 **illustrates** the analog wave that is produced by the function **sin(x),** where the **magnitude** of this **analog** wave is 2. The **value** of 2sin(x) is 0 at** x equal** to 0.

**Figure 3 ** illustrates the sine wave produced by 2sin(x) and -2sin(x).

## Numerical Problem of an Analog Function

The **analog** wave of **2sin(x)** and **2cos(x)** is shown below. You are required to find the following parts.

a) Find the value of **2sin(x)** when the **value** of x is **2 and 0**.

b) Find the value of **2sin(x)** when the **value** of x is **4 and 0**.

c) Find the value of **2sin(x)** when the value of x is **6 and 0**.

d) Find the value of **2sin(x)** when the **value** of x is **8 and 0**.

e) Find the value of **2cos(x)** when the **value** of x is **2 and 0**.

f) Find the value of **2cos(x)** when the **value** of x is **4 and 0**.

g) Find the value of **2cos(x)** when the **value** of x is **6 and 0**.

h) Find the value of **2cos(x)** when the **value** of x is **8 and 0**.

### Solution

**a) **We have to find the **value** of **2sin(x)** at x equal to 2 and 0.

So,

**= 2sin(x)**

At x **equal** to 2 **results** in:

**= 2sin(2)**

**= 2 x 0.90929742682**

**= 1.81859485365**

Now at x **equal** to 0 results in:

**= 2sin(0)**

**= 2 x 0**

**= 0**

Thus the value of **2sin(2)** is **1.81859485365 and **the **value** is 0 at **2sin(0).**

**b) **We have to find the **value** of **2sin(x)** at x equal to 4 and 0.

So,

**= 2sin(x)**

At x **equal** to 4 **results** in:

**= 2sin(4)**

**= 2 x (- 0.7568024953 )**

**= – 1.51360499062**

Now at x **equal** to 0 **results** in:

**= 2sin(0)**

**= 2 x 0**

**= 0**

Thus the value of **2sin(4)** is **– 1.51360499062**** and **the **value** is 0 at **2sin(0).**

**c) **We have to **find** the value of **2sin(x)** at x **equal** to 6 and 0.

So,

**= 2sin(x)**

At x **equal** to 4 **results** in:

**= 2sin(6)**

**= 2 x (- 0.27941549819 )**

**= – 0.55883099639**

Now at x **equal** to 0 **results** in:

**= 2sin(0)**

**= 2 x 0**

**= 0**

Thus the value of **2sin(6)** is **-0.55883099639 and **the **value** is 0 at **2sin(0).**

**d) **We have to find the **value** of **2sin(x)** at x **equal** to 8 and 0.

So,

**= 2sin(x)**

At x **equal** to 4 **results** in:

**= 2sin(8)**

**= 2 x (0.98935824662 )**

**= 1.97871649325**

Now at x **equal** to 0 **results** in:

**= 2sin(0)**

**= 2 x 0**

**= 0**

Thus the value of **2sin(8)** is 1.97871649325** and **the **value** is 0 at **2sin(0).**

**e) **We have to find the **value** of **2cos(x)** at x equal to **2 and 0**.

So,

**= 2cos(x)**

At x **equal** to 2 **results** in:

**= 2cos(2)**

**= 2 x (-0.41614683654)**

**= -0.83229367309**

Now at x **equal** to 0 **results** in:

**= 2cos(0)**

**= 2 x 1**

**= 2**

Thus the value of **2cos(2)** is **-0.83229367309**** and **the **value** is 1 at **2cos(0).**

**f) **We have to find the **value** of **2cos(x)** at x equal to **4** and **0**.

So,

**= 2cos(x)**

At x **equal** to 4 **results** in:

**= 2cos(4)**

**= 2 x (-0.65364362086)**

**= -1.30728724173**

Now at x **equal** to 0 **results** in:

**= 2cos(0)**

**= 2 x 1**

**= 2**

Thus the value of **2cos(4)** is **-1.30728724173 and **the **value** is 1 at **2cos(0).**

**g) **We have to **find** the value of **2cos(x)** at x equal to **6** and **0**.

So,

**= 2cos(x)**

At x **equal** to 6 **results** in:

**= 2cos(6)**

**= 2 x (0.96017028665)**

**= 1.9203405733**

Now at x **equal** to 0 **results** in:

**= 2cos(0)**

**= 2 x 1**

**= 2**

Thus the value of **2cos(6)** is **1.9203405733 and **the **value** is 1 at **2cos(0).**

**h) **We have to **find** the value of **2cos(x)** at x equal to **8 and 0**.

So,

**= 2cos(x)**

At x **equal** to 8 **results** in:

**= 2cos(8)**

**= 2 x (-0.1455000338)**

**= -0.29100006761**

Now at x **equal** to 0 **results** in:

**= 2cos(0)**

**= 2 x 1**

**= 2**

Thus the value of **2cos(8)** is **-0.29100006761 and **the **value** is 1 at **2cos(0).**

*All mathematical drawings and images were created with GeoGebra.*